Binomial Theorem JEE Main Questions 2027 with Solutions

Weightage of Binomial Theorem in JEE Main 2027

The analysis of JEE Main 2026 shows that the Binomial Theorem chapter remains the important chapter in the JEE Main Maths exam. The table below shows the session-wise number of questions, marks, and weightage of the chapter.

Binomial Theorem JEE Main Questions 2027

Question 1: For an integer $\mathrm{n} \geq 2$, if the arithmetic mean of all coefficients in the binomial expansion of $(x+y)^{2 n-3}$ is 16 , then the distance of the point $P\left(2 n-1, n^2-4 n\right)$ from the line $x+y=8$ is:

1) $\sqrt{2}$
2) $2 \sqrt{2}$
3) $5 \sqrt{2}$
4) $3 \sqrt{2}$

Correct answer: (4)

Solution:

No. of terms in $(x+y)^{(2 n-3)} \Rightarrow\left[\begin{array}{c}(2 n-3+1) \\ (2 n-2)\end{array}\right.$
$\therefore$ sum of all coefficients $=2^{2 n-3}$
(Put $\mathrm{x}=\mathrm{y}=1$ )
$\therefore$ Arithmetic mean of all coefficients

$\begin{aligned} & =\left(\frac{2^{2 n-3}}{2 n-2}\right)=16 \\ & \Rightarrow 2^{2 n-3}=2^5(n-1) \Rightarrow n=5 \\ & \therefore P\left(2 n-1, n^2-4 n\right)=(9,5)\end{aligned}$

$\begin{aligned} & \mathrm{x}+\mathrm{y}=8 \\ & \therefore \mathrm{PM}=\left|\frac{9+5-8}{\sqrt{2}}\right|=\frac{6}{\sqrt{2}}=\frac{3 \times 2}{\sqrt{2}}=3 \sqrt{2}\end{aligned}$

Hence, the correct answer is option (4).

Question 2: The remainder when $\left((64)^{(64)}\right)^{(64)}$ is divided by 7 is equal to

1) 4
2) 1
3) 3
4) 6

Correct answer: (2)

Solution:

Let $\mathrm{N}=\left((64)^{64}\right)^{64}$

$\begin{aligned}
& \mathbf{N}=(64)^{64^2} \\
& \mathbf{N}=(1+63)^{64^2}, \text { let } 64^2=\mathrm{n}
\end{aligned}$

Expanding by binomial

$\begin{aligned}
& \mathrm{N}=(1+63)^{\mathrm{n}}=1+{ }^{\mathrm{n}} \mathrm{C}_1 63+{ }^{\mathrm{n}} \mathrm{C}_2(63)^2+\ldots \ldots \\
& =1+63 \lambda=1+7(9 \lambda)
\end{aligned}$

Remainder when divided by 7 is 1.

Hence, the answer is option (2).

Question 3: If $\sum_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}+3}{2^r}\right) \cdot{ }^9 \mathrm{C}_{\mathrm{r}}=\alpha\left(\frac{3}{2}\right)^9-\beta, \quad \alpha, \beta \in \mathrm{N}, \quad$ then $(\alpha+\beta)^2$ is equal to :

1) 27
2) 9
3) 81
4) 18

Correct answer: (3)

Solution:

Given that

$\sum_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}_{\mathrm{r}}=\alpha\left(\frac{3}{2}\right)^9-\beta, \alpha, \beta \in \mathrm{N}$

Now, use binomials,

$\begin{aligned}
& \sum_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}_{\mathrm{r}}=\sum_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}_{\mathrm{r}}+\sum_{\mathrm{r}=1}^9\left(\frac{3}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}_{\mathrm{r}} \\
& =\sum_{\mathrm{r}=1}^9\left(\frac{9}{2^{\mathrm{r}}}\right) \cdot{ }^8 \mathrm{C}_{\mathrm{r}-1}+3 \sum_{\mathrm{r}=1}^9{ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}\left[\mathrm{U} \sin \mathrm{~g} \frac{{ }^9 \mathrm{C}_{\mathrm{r}}}{{ }^8 \mathrm{C}_{\mathrm{r}-1}}=\frac{9}{\mathrm{r}}\right] \\
& =\frac{9}{2} \sum_{\mathrm{r}=1}^9{ }^8 \mathrm{C}_{\mathrm{r}-1}\left(\frac{1}{2}\right)^{\mathrm{r}-1}+3\left(\sum_{\mathrm{r}=0}^9\left({ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}\right)-1\right) \\
& =\frac{9}{2}\left(1+\frac{1}{2}\right)^8+3\left(\left(1+\frac{1}{2}\right)^9-1\right) \\
& =\frac{9}{2} \cdot\left(\frac{3}{2}\right)^8+3\left(\frac{3}{2}\right)^9-3=6 \cdot\left(\frac{3}{2}\right)^9-3
\end{aligned}$

On comparing it with, $\alpha\left(\frac{3}{2}\right)^9-\beta, \alpha, \beta \in \mathrm{N}$

We have, $\alpha=6, \beta=3$
Thus $(\alpha+\beta)^2$

$(6+3)^2$

$9^2=81$

Hence, the answer is option (3).

Question 4: The sum of all rational terms in the expansion of $\left(1+2^{1 / 3}+3^{1 / 2}\right)^6$ is equal to____________.

1) 612
2) -
3) -
4) -

Correct answer: (1)

Solution:

Consider the expansion of:

$
\left(1 + 2^{\frac{1}{3}} + 3^{\frac{1}{2}}\right)^6
$

Using the multinomial theorem, the general term is:

$
\binom{6}{r_1, r_2, r_3} (1)^{r_1} \left(2^{\frac{1}{3}}\right)^{r_2} \left(3^{\frac{1}{2}}\right)^{r_3}
$

where

$
r_1 + r_2 + r_3 = 6
$

and

$
\binom{6}{r_1, r_2, r_3} = \frac{6!}{r_1! \, r_2! \, r_3!}
$

Now, summing the terms where the powers yield integers:

$
\text{Sum} = \binom{6}{6,0,0} (1)^6 + \binom{6}{4,0,2} (2^{\frac{1}{3} \times 0}) (3^{\frac{1}{2} \times 2}) + \binom{6}{2,0,4} (3)^{2} + \binom{6}{0,0,6} (3)^{3}
$

$
+ \binom{6}{3,3,0} (2)^{1} + \binom{6}{1,3,2} (2)^1 (3)^1 + \binom{6}{0,6,0} (2)^2
$

Evaluating the binomial coefficients and powers:

$
= 1 + 45 + 135 + 27 + 40 + 360 + 4 = 612
$

Hence, the answer is 612.

Question 5: If in the expansion of $(1+x)^p(1-x)^q$, the coefficients of $x$ and $x^2$ are 1 and -2 , respectively, then $\mathrm{p}^2+\mathrm{q}^2$ is equal to:

1) 8
2) 18
3) 13
4) 20

Correct answer: (3)

Solution:

We have:

$
(1+x)^p(1-x)^q = \left({}^p C_0 + {}^p C_1 x + {}^p C_2 x^2 + \ldots\right)\left({}^q C_0 - {}^q C_1 x + {}^q C_2 x^2 - \ldots\right)
$

To find the coefficient of $x$, we consider terms whose degrees add up to 1:

Coefficient of $x = {}^p C_0 \cdot ( - {}^q C_1 ) + {}^p C_1 \cdot {}^q C_0 = -{}^q C_1 + {}^p C_1 = 1
$

Given that $p - q = 1$, so:

$
p = q + 1
$

Now for the coefficient of $x^2$, the relevant terms are:

Coefficient of $x^2 = {}^p C_0 \cdot {}^q C_2 - {}^p C_1 \cdot {}^q C_1 + {}^p C_2 \cdot {}^q C_0 = -2
$

Now express binomial coefficients using formulas:

$
{}^q C_2 = \frac{q(q - 1)}{2}, \quad {}^p C_2 = \frac{p(p - 1)}{2}, \quad {}^p C_1 = p, \quad {}^q C_1 = q
$

Substitute these into the expression:

$
\frac{q(q - 1)}{2} - pq + \frac{p(p - 1)}{2} = -2
$

Multiply the entire equation by 2 to eliminate denominators:

$
q(q - 1) - 2pq + p(p - 1) = -4
$

Simplify:

$
q^2 - q - 2pq + p^2 - p = -4
$

Now substitute $p = q + 1$ into the equation:

$
q^2 - q - 2q(q + 1) + (q + 1)^2 - (q + 1) = -4
$

Expand:

$
q^2 - q - 2q^2 - 2q + q^2 + 2q + 1 - q - 1 = -4
$

Simplify:

$
(q^2 - 2q^2 + q^2) + (-q - 2q + 2q - q) + (1 - 1) = -4
\Rightarrow 0 - 2q = -4 \Rightarrow q = 2
$

Now:

$
p = q + 1 = 3
$

Therefore:

$
p^2 + q^2 = 3^2 + 2^2 = 9 + 4 = 13
$

Hence, the correct answer is option (3).

Question 6: Suppose A and B are the coefficients of $30^{\text {th }}$ and $12^{\text {th }}$ terms respectively in the binomial expansion of $(1+x)^{2 n-1}$. If $2 A=5 B$, then $n$ is equal to:

1) 22
2) 21
3) 20
4) 19

Correct answer: (2)

Solution:

We are given:

$
A = \binom{2n - 1}{29}, \quad B = \binom{2n - 1}{11}
$

and the relation:

$
2A = 5B
$

Step 1: Write the equation using factorial definition of combinations

$
2 \binom{2n - 1}{29} = 5 \binom{2n - 1}{11}
$

$
2 \cdot \frac{(2n - 1)!}{29! (2n - 30)!} = 5 \cdot \frac{(2n - 1)!}{11! (2n - 12)!}
$

Step 2: Cancel common factorial terms

$
2 \cdot \frac{1}{29! (2n - 30)!} = 5 \cdot \frac{1}{11! (2n - 12)!}
$

Step 3: Rearrange

$
\frac{2}{29! (2n - 30)!} = \frac{5}{11! (2n - 12)!}
$

Cross-multiplied:

$
2 \cdot 11! (2n - 12)! = 5 \cdot 29! (2n - 30)!
$

Step 4: Express factorial terms to cancel

Note:

$
29! = 29 \times 28 \times \cdots \times 12 \times 11!
$

and

$
(2n - 12)! = (2n - 12)(2n - 13) \cdots (2n - 29) (2n - 30)!
$

So substituting back:

$
2 \cdot 11! \cdot (2n - 12)(2n - 13) \cdots (2n - 29) (2n - 30)! = 5 \cdot 29 \times 28 \times \cdots \times 12 \times 11! \cdot (2n - 30)!
$

Step 5: Cancel common terms $11!$ and $(2n - 30)!$

$
2 \cdot (2n - 12)(2n - 13) \cdots (2n - 29) = 5 \cdot 29 \times 28 \times \cdots \times 12
$

Rewrite the right side as:

$
5 \times (29 \times 28 \times \cdots \times 12)
$

and the left side as:

$
2 \times \prod_{k=12}^{29} (2n - k)
$

Step 6: Observe that the products have the same number of terms (from 12 to 29, total 18 terms). For the equality to hold, compare term by term:

This leads to the key relation:

$
2n - 12 = 30
$

because the smallest term on the left side equals the smallest term on the right side plus 1.

Step 7: Solve for $n$

$
2n - 12 = 30 \implies 2n = 42 \implies n = 21
$

Hence, the correct answer is option (2).

Question 7: The remainder, when $7^{103}$ is divided by 23, is equal to:

Binomial Theorem JEE Main Questions - Binomial Theorem JEE Main Questions - 1) 14
2) 9
3) 17
4) 6

Correct answer: (1)

Solution:

We start by expressing $7^{103}$ as:

$
7^{103} = 7 \times 7^{102} = 7 \times (7^3)^{34}
$

Note that

$
7^3 = 343 = 345 - 2
$

Since $345 = 23 \times 15$, it follows that $343 \equiv -2 \pmod{23}$.

Thus,

$
7^{103} \equiv 7 \times (-2)^{34} \pmod{23}.
$

Since $(-2)^{34} = 2^{34}$ (because 34 is even),

$
7^{103} \equiv 7 \times 2^{34} \pmod{23}.
$

Now, compute $2^{34} \pmod{23}$:

First, note that $2^8 = 256 \equiv 3 \pmod{23}$, since $256 - 11 \times 23 = 256 - 253 = 3$.

Therefore,

$
2^{32} = (2^8)^4 \equiv 3^4 = 81 \pmod{23}.
$

Now, $81 \equiv 81 - 3 \times 23 = 81 - 69 = 12 \pmod{23}$.

Hence,

$
2^{34} = 2^{32} \times 2^2 \equiv 12 \times 4 = 48 \equiv 48 - 2 \times 23 = 48 - 46 = 2 \pmod{23}.
$

Putting it all together:

$
7^{103} \equiv 7 \times 2 = 14 \pmod{23}.
$

The remainder is $14$.

Hence, the correct answer is option (1).

Question 8: The sum of all rational terms in the expansion of $(2+\sqrt{3})^8$ is

1) 16923 
2) 3763
3) 33845 
4) 18817

Correct answer: (4)

Solution:

Let $S = (2 + \sqrt{3})^8$

To find the sum of all rational terms, we use the binomial expansion:

$(2 + \sqrt{3})^8 = \sum_{k=0}^{8} \binom{8}{k} \cdot 2^{8-k} \cdot (\sqrt{3})^k$

Only even powers of $\sqrt{3}$ will give rational terms. So we take even $k$:
$k = 0, 2, 4, 6, 8$

Now calculate:

$\binom{8}{0} \cdot 2^8 = 1 \cdot 256 = 256$

$\binom{8}{2} \cdot 2^6 \cdot (\sqrt{3})^2 = 28 \cdot 64 \cdot 3 = 5376$

$\binom{8}{4} \cdot 2^4 \cdot (\sqrt{3})^4 = 70 \cdot 16 \cdot 9 = 10080$

$\binom{8}{6} \cdot 2^2 \cdot (\sqrt{3})^6 = 28 \cdot 4 \cdot 27 = 3024$

$\binom{8}{8} \cdot (\sqrt{3})^8 = 1 \cdot 81 = 81$

Now add all rational terms:

$256 + 5376 + 10080 + 3024 + 81 = 18817$

Hence, the answer is option (4).

Question 9: The number of integral terms in the expansion of $\left(5^{\frac{1}{2}}+7^{\frac{1}{8}}\right)^{1016}$ is

1) 127
2) 130
3) 129
4) 128

Correct answer: (4)

Solution:

$\begin{aligned} & T_{r+1}=^{1016}C_r \cdot\left(7^{\frac{r}{8}}\right)^4 \cdot 5^{\frac{1}{2}(1016-r)} \\ & \Rightarrow \frac{r}{8} \text { and } \frac{1016-r}{2} \in \operatorname{Integer} \\ & \Rightarrow 8 / r \Rightarrow r=0,8,16, \ldots 1016 \\ & \Rightarrow r=0 \times 8,1 \times 8, \ldots 127 \times 8 \\ & \Rightarrow \text { Total } 128  \text { such that } T_{r+1} \text { is rational }\end{aligned}$

Hence, the answer is option (4).

Question 10: In the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n, \mathrm{n} \in \mathrm{N}$, if the ratio of $15^{\text {th }}$ term from the beginning to the $15^{\text {th }}$ term from the end is $\frac{1}{6}$, then the value of ${ }^n C_3$ is:

1) 4060
2) 1040
3) 2300
4) 4960

Correct answer: (3)

Solution:

$\begin{aligned}
& T_{r+1}={ }^n C_r\left(2^{1 / 3}\right)^{n-r}\left(\frac{1}{3^{1 / 3}}\right)^r \\
& r=14 \\
& T_{15}={ }^n C_{14}\left(2^{1 / 3}\right)^{n-14}\left(\frac{1}{3^{1 / 3}}\right)^{14}
\end{aligned}$

$\mathrm{T}_{15}^{\prime}=15^{\text {th }}$ term from last is $(\mathrm{n}-13)^{\text {th }}$ term from beginning.

$\begin{aligned}
& T_{15}^{\prime}={ }^n C_{n-14}\left(2^{1 / 3}\right)^{14}\left(\frac{1}{3^{1 / 3}}\right)^{n-14} \\
& \Rightarrow \frac{T_{15}}{T_{15}^{\prime}}=\frac{{ }^n C_{14}\left(2^{1 / 3}\right)^{n-14}\left(\frac{1}{3^{1 / 3}}\right)^{14}}{{ }^n C_{n-14}\left(2^{1 / 3}\right)^{14}\left(\frac{1}{3^{1 / 3}}\right)^{n-14}}=\frac{1}{6} \\
& =\left(2^{1 / 3}\right)^{n-28}\left(3^{1 / 3}\right)^{n-28}=\frac{1}{6} \\
& \Rightarrow6^{\frac{n-28}{3}}=6^{-1}
\end{aligned}$

For same bases, exponents will be same,so, we can write,

$\frac{n-28}{3}=-1$

$n-28=-3$

$n=25$

So, ${ }^{\mathrm{n}} \mathrm{C}_3={ }^{25} \mathrm{C}_3=2300$

Hence, the correct answer is option (3).

Question 11: If $1^2 \cdot\left({ }^{15} \mathrm{C}_1\right)+2^2 \cdot\left({ }^{15} \mathrm{C}_2\right)+3^2 \cdot\left({ }^{15} \mathrm{C}_3\right)+\ldots .+15^2 \cdot\left({ }^{15} \mathrm{C}_{15}\right)=$ $2^{\mathrm{m}} \cdot 3^{\mathrm{n}} \cdot 5^{\mathrm{k}}$, where $\mathrm{m}, \mathrm{n}, \mathrm{k} \in \mathrm{N}$, then $\mathrm{m}+\mathrm{n}+\mathrm{k}$ is equal to :-

1) 19
2) 21
3) 18
4) 20

Correct answer: (1)

Solution:

We have the sum,

$\begin{aligned} & \sum_{\mathrm{r}=1}^{15} \mathrm{r}^2\left({ }^{15} \mathrm{C}_{\mathrm{r}}\right) \Rightarrow 15 \sum_{\mathrm{r}=1}^{15} \mathrm{r}^{14} \mathrm{C}_{\mathrm{r}-1} \\ &= 15 \sum_{\mathrm{r}=1}^{15}(\mathrm{r}-1+1){ }^{14} \mathrm{C}_{\mathrm{r}-1} \\ &= 15 \cdot 14 \sum_{\mathrm{r}=1}^{15}{ }^{13} \mathrm{C}_{\mathrm{r}-2}+15 \sum_{\mathrm{r}=1}^{15}{ }^{14} \mathrm{C}_{\mathrm{r}-1} \\ &= 15 \cdot 14 \cdot 2^{13}+15 \cdot 2^{14} \\ & =3^1 \cdot 2^{13}(70+10) \\ & =3^1 \cdot 2^{13} \cdot 80 \\ &= 3^1 \cdot 5^1 \cdot 2^{17} \\ & \mathrm{~m}=17, \quad \mathrm{n}=1, \quad \mathrm{k}=1\end{aligned}$

Hence, the answer is option (1).

Question 12: Let $\left(1+x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots .+a_{20} x^{20}$. If $\left(a_1+a_3+a_5+\ldots .+a_{19}\right)-11 \mathrm{a}_2=121 \mathrm{k}$, then k is equal to _____.

1) 239
2) ---
3) ---
4) ---

Correct answer: (1)

Solution:

Let $f(x)=\left(1+x+x^2\right)^{10}=\sum_{r=0}^{20} a_r x^r$
The sum of odd coefficients: $S_{\text {odd }}=a_1+a_3+a_5+\cdots$ $+a_{19}$

Subtracting $11 a_2$ from above will give the answer

$\begin{aligned}
& S_{\text {odd }}=\frac{f(1)-f(-1)}{2} \\
& f(1)=(1+1+1)^{10}=3^{10} \\
& f(-1)=(1-1+1)^{10}=(1)^{10}=1 \\
& S_{\text {odd }}=\sum_{\text {odd } r} a_r=\frac{3^{10}-1}{2}
\end{aligned}$

Now for $a_2$

$1+x+x^2=\frac{1-x^3}{1-x} \Rightarrow f(x)=\left(\frac{1-x^3}{1-x}\right)=\frac{\left(1-x^3\right)^{10}}{(1-x)^{10}}$

Now use:

$\begin{aligned}
& \left(1-x^3\right)^{10}=\sum_{x=0}^{10}(-1)^k \times ^{10}C_{k} x^{3 k} \\
& (1-x)^{-10}=\sum_{r=0}^{\infty}{^{r+9}C_{9}} x^r
\end{aligned}$

So

$f(x)=\left(\sum_{k=0}^{10}(-1)^k\times ^{10}C_{k} x^{3 k}\right) \cdot\left(\sum_{r=0}^{\infty}{^{r+9}C_{9}} x^r\right)$

Only the term with $x^0$ from the first sum (i.e., $k=0$ ) can contribute to $x^2$, since all other $k \geq 1$ gives $x^{3 k} \geq$ $x^3$

From $\left(1-x^3\right)^{10}$ : the $x^0$ term is $^{10}C_{0}=1$
From $(1-x)^{-10}:$ the coefficient of $x^2$ is

$^{2+9}C_{9}=^{11}C_{9}=55$

Hence, $a_2=1.55=55$

Now, $S_{odd }-11 a_2=\frac{3^{10}-1}{2}-11 \cdot 55=121 k$
$3^{10}=59049$

So:

$S=\frac{59049-1}{2}-605=\frac{59048}{2}-605$

$=29524-605=28919$

So:

$121 k=28919 \Rightarrow k=\frac{28919}{121}=239$
Hence, the answer is 239.

Question 13: Let m and $\mathrm{n},(\mathrm{m}<\mathrm{n})$ be two 2-digit numbers. Then the total numbers of pairs $(\mathrm{m}, \mathrm{n})$, such that $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=6$, is ______

1) 64
2) -
3) -
4) -

Correct answer: (1)

Solution:

Let $\mathrm{m}=6 \mathrm{a}, \mathrm{n}=6 \mathrm{~b}$

$\mathrm{m}<\mathrm{n} \Rightarrow \mathrm{a}<\mathrm{b}$

where $\mathrm{a}$ &$\mathrm{~b}$ are co-prime numbers also since $m$ & $n$ are 2 digit numbers, so

$10 \leq \mathrm{m} \leq 99$& $10 \leq \mathrm{n} \leq 99$

i.e. $2 \leq a \leq 16 \& 2 \leq b \leq 16$
( $\because$ a is integer)

Now
$2 \leq \mathrm{a}<\mathrm{b} \leq 16$ & $\quad \mathrm{a} \& \mathrm{~b}$ are co-prime

$\begin{aligned}\text{If }
& a=2, b=3,5,7,9,11,13,15 \\
& a=3, b=4,5,7,8,10,11,13,14,16 \\
& a=4, b=5,7,9,11,13,15 \\
& a=5, b=6,7,8,9,11,12,13,14,16 \\
& a=6, b=7,11,13 \\
& a=7, b=8,9,10,11,12,13,15,16 \\
& a=8, b=9,11,13,15 \\
& a=9, b=10,11,13,14,16 \\
& a=10, b=11,13 \\
& a=11, b=12,13,14,15,16 \\
& a=12, b=13
\end{aligned}$

$\begin{aligned}
& a=13, b=14,15,16 \\
& a=14, b=15 \\
& a=15, b=16
\end{aligned}$

64 ordered pairs

Hence, the answer is (64).

Question 14: Let $\alpha, \beta, \gamma$ and $\delta$ be the coefficients of $x^7, x^5, x^3$ and $x$ respectively in the expansion of $\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5, x>1$. If $u$ and $v$ satisfy the equations
$
\begin{aligned}
& \alpha u+\beta v=18 \\
& \gamma u+\delta v=20
\end{aligned}
$
then $u+v$ equals:

1) 5
2) 4
3) 3
4) 8

Correct answer: (1)

Solution:

We have the binomial sum,
$
\begin{aligned}
& \left(\mathrm{x}+\sqrt{\mathrm{x}^3-1}\right)^5+\left(\mathrm{x}-\sqrt{\mathrm{x}^3-1}\right)^5 \\
& =2\left\{{ }^5 \mathrm{C}_0 \cdot \mathrm{x}^5+{ }^5 \mathrm{C}_2 \cdot \mathrm{x}^3\left(\mathrm{x}^3-1\right)+{ }^5 \mathrm{C}_4 \cdot \mathrm{x}\left(\mathrm{x}^3-1\right)^2\right\} \\
& =2\left\{5 \mathrm{x}^7+10 \mathrm{x}^6+\mathrm{x}^5-10 \mathrm{x}^4-10 \mathrm{x}^3+5 \mathrm{x}\right\} 
\end{aligned}
$
Now,
$\begin{aligned} & \alpha=\text { coefficient of } x^7=10 \\ & \beta=\text { coefficient of } x^5=2 \\ & \gamma=\text { coefficient of } x^3=-20 \\ & \delta=\text { coefficient of } x=10\end{aligned}$
Now, $10 \mathrm{u}+2 \mathrm{v}=18---(1)$
and $-20 u+10 v=20----(2)$
Solving these equations, we get,
$u=1, v=4$
$\therefore u+v=1+4=5$
Hence, the correct answer is option (1).

Question 15: Let the coefficients of three consecutive terms $T_r$, $T_{r+1}$, and $T_{r+2}$ in the binomial expansion of $(a+b)^{12}$ be in a G.P. and let $p$ be the number of all possible values of $r$. Let $q$ be the sum of all rational terms in the binomial expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{12}$. Then $\mathrm{p}+\mathrm{q}$ is equal to:

1) 283
2) 295
3) 287
4) 299

Correct answer: (1)

Solution:

Given:

Let the coefficients of three consecutive terms $T_r$, $T_{r+1}$, and $T_{r+2}$ in the binomial expansion of $(a + b)^{12}$ be in a geometric progression (G.P.).

Let $p$ be the number of all possible values of $r$.

Let $q$ be the sum of all rational terms in the binomial expansion of $\left( \sqrt[4]{3} + \sqrt[3]{4} \right)^{12}$.

To find: $p + q$ let us find $p$

In the binomial expansion of $(a + b)^{12}$, the general term is:
$
T_{r+1} = \binom{12}{r} a^{12 - r} b^r
$
We are only concerned with the coefficients: $\binom{12}{r - 1}$, $\binom{12}{r}$, $\binom{12}{r + 1}$.

Let these be in G.P. Then:
$
\frac{\binom{12}{r}}{\binom{12}{r - 1}} = \frac{\binom{12}{r + 1}}{\binom{12}{r}}
$

Cross-multiplying:
$
\left( \binom{12}{r} \right)^2 = \binom{12}{r - 1} \cdot \binom{12}{r + 1}
$

We use the identity:
$
\frac{\binom{n}{r}}{\binom{n}{r - 1}} = \frac{n - r + 1}{r}
$

Apply to both sides:

LHS:
$
\frac{\binom{12}{r}}{\binom{12}{r - 1}} = \frac{13 - r}{r}
$

RHS:
$
\frac{\binom{12}{r + 1}}{\binom{12}{r}} = \frac{12 - r}{r + 1}
$

Equating:
$
\frac{13 - r}{r} = \frac{12 - r}{r + 1}
$

Cross-multiplying:
$
(13 - r)(r + 1) = (12 - r)(r)
$

Expand both sides:

LHS:
$
(13 - r)(r + 1) = 13r + 13 - r^2 - r = -r^2 + 12r + 13
$

RHS:
$
(12 - r)(r) = 12r - r^2
$

Set equal:
$
-r^2 + 12r + 13 = 12r - r^2
$

Simplify:
$
-r^2 + 12r + 13 = 12r - r^2
\Rightarrow \text{Both sides are equal only if } 13 = 0,
$

Which is a contradiction.

So, no such $r$ exists. Hence, $p = 0$

Now find $q$

Now, consider:
$
\left( \sqrt[4]{3} + \sqrt[3]{4} \right)^{12}
= \sum_{r=0}^{12} \binom{12}{r} \left(3^{\frac{1}{4}}\right)^{12 - r} \left(4^{\frac{1}{3}}\right)^r
$

Each term is:
$
\binom{12}{r} \cdot 3^{\frac{12 - r}{4}} \cdot 4^{\frac{r}{3}}
$

This term is rational only if the exponents of 3 and 4 are integers.

That is, both:
$\frac{12 - r}{4} \in \mathbb{Z}$
$\frac{r}{3} \in \mathbb{Z}$

So:
$12 - r \equiv 0 \pmod{4} \Rightarrow r \equiv 0 \pmod{4}$
$r \equiv 0 \pmod{3}$

Thus, $r$ must be divisible by both 3 and 4 ⇒ $\mathrm{LCM}(3, 4) = 12$

So only possible values of $r$ in $0 \leq r \leq 12$ are: $r = 0$ and $r = 12$

Let’s evaluate those two terms:

When $r = 0$:

$
T_1 = \binom{12}{0} \cdot 3^{12/4} \cdot 4^0 = 1 \cdot 3^3 \cdot 1 = 27
$

When $r = 12$:

$
T_{13} = \binom{12}{12} \cdot 3^0 \cdot 4^{12/3} = 1 \cdot 1 \cdot 4^4 = 256
$

So total sum of rational terms $q = 27 + 256 = 283$

Therefore, $p + q = 0 + 283 = 283$

Hence, the correct answer is option (1).

Question 16: The least value of n for which the number of integral terms in the Binomial expansion of $(\sqrt[3]{7}+\sqrt[12]{11})^{\mathrm{n}}$ is 183, is:

1) 2184
2) 2148
3) 2172
4) 2196

Correct answer: (1)

Solution:

We are given the binomial expansion:
$
\left(\sqrt[3]{7} + \sqrt[12]{11}\right)^n = \left(7^{\frac{1}{3}} + 11^{\frac{1}{12}}\right)^n.
$

The general term $ T_{r+1} $ is:
$
T_{r+1} = \binom{n}{r} \left(7^{\frac{1}{3}}\right)^{n-r} \left(11^{\frac{1}{12}}\right)^r = \binom{n}{r} 7^{\frac{n-r}{3}} 11^{\frac{r}{12}}.
$

Condition for integral terms

For $T_{r+1}$ to be an integer (integral term), the exponents of 7 and 11 must be integers.

- The exponent of 7 is $\frac{n - r}{3}$, so $n - r$ must be divisible by 3.
- The exponent of 11 is $\frac{r}{12}$, so $r$ must be divisible by 12.

Since $r$ varies from 0 to $n$, the divisibility by 12 restricts $r$ to values:
$
r = 12k, \quad k \in \mathbb{W}, \quad 0 \leq 12k \leq n.
$

Number of integral terms equals the number of valid $r$:
$
\text{Number of integral terms} = [ \frac{n}{12}] + 1.
$

Given the number of integral terms is 183, so:
$
[\frac{n}{12}] + 1 = 183 \implies [\frac{n}{12}] = 182.
$

Thus,
$
n \geq 12 \times 182 = 2184.
$

The least such $n$ is: $2184$

Hence, the correct answer is option (1).

Question 17: The term independent of $x$ in the expansion of  $\left(\frac{(x+1)}{\left(x^{2 / 3}+1-x^{1 / 3}\right)}-\frac{(x-1)}{\left(x-x^{1 / 2}\right)}\right)^{10}, x>1$ is:

1) 210
2) 150
3) 240
4) 120

Correct answer: (1)

Solution:

Simplify each term:

$\left(\frac{(x+1)}{\left(x^{\frac{2}{3}}+1-x^{\frac{1}{3}}\right)}-\frac{(x-1)}{\left(x-x^{\frac{1}{2}}\right)}\right)^{10}$

Use formula of $a^3+b^3=(a+b)(a^2+b^2-ab)$ and cancel the common terms and rationalize the second term,

$\Rightarrow \left(\frac{(x^{1/3}+1)(x^{2/3}+1-x^{1/3})}{\left(x^{2/3}+1-x^{1/3}\right)}-\frac{(x-1)}{\left(x-x^{\frac{1}{2}}\right)}\times \frac{x+\sqrt{x}}{x+\sqrt{x}}\right)^{10}$

$=\left(x^{\frac{1}{3}}+1-\left(\frac{\sqrt{x}+1}{\sqrt{x}}\right)\right)^{10}$

$=\left(x^{\frac{1}{3}}-\frac{1}{\sqrt{x}}\right)^{10}$

Now, use the formula of general term,

$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{\frac{10-\mathrm{r}}{3}}(-1)^{\mathrm{r}}(\mathrm{x})^{-\frac{\mathrm{r}}{2}}$

$\frac{10-r}{3}-\frac{r}{2}=0$

$(20-2 r)-3 r=0$

$r=4$

Substitute this value into the independent term,

$\Rightarrow{ }^{10} \mathrm{C}_{4}(-1)^{4}=210$

Hence, the correct answer is option (1).

Question 18: If $\displaystyle\sum_{\mathrm{r}=1}^{30} \frac{\mathrm{r}^2\left({ }^{30} \mathrm{C}_{\mathrm{r}}\right)^2}{{ }^{30} \mathrm{C}_{\mathrm{r}-1}}=\alpha \times 2^{29}$, then $\alpha$ is equal to______________.

1) 465
2) -
3) -
4) -

Correct answer: (1)

Solution:

Given,$\displaystyle\sum_{r=1}^{30} \frac{r^2 \binom{30}{r}^2}{\binom{30}{r-1}}$

Simplify the ratio of binomial coefficients:
$
\frac{\binom{30}{r}}{\binom{30}{r-1}} = \frac{30 - r + 1}{r} = \frac{31 - r}{r}
$

Rewrite the sum as:
$
\displaystyle\sum_{r=1}^{30} r^2 \left(\frac{31 - r}{r}\right) \binom{30}{r} $

$= \displaystyle\sum_{r=1}^{30} (31 - r) r \binom{30}{r}$

$
=\displaystyle\sum_{r=1}^{30} (31 - r) r \frac{30!}{r! (30 - r)!} $               ($\binom{30}{r} = \frac{30!}{r! (30 - r)!}$)

$= \displaystyle\sum_{r=1}^{30} (31 - r) \frac{30!}{(r-1)! (30 - r)!}$                          

$
= 30 \displaystyle\sum_{r=1}^{30} (31 - r) \frac{29!}{(r-1)! (30 - r)!}
$

Recognize the binomial coefficient $\binom{29}{k}$:
$
\frac{29!}{k! (29 - k)!} = \binom{29}{k}
$

By substituting $k = 30 - r$, the sum becomes:
$
30 \displaystyle\sum_{r=1}^{30} (31 - r) \binom{29}{30 - r}
$

Split the sum into two parts:
$
30 \left( \displaystyle\sum_{r=1}^{30} (31 - r) \binom{29}{30 - r} \right) = 30 \left( \displaystyle\sum_{k=0}^{29} (k+1) \binom{29}{k} \right)
$

Use the standard binomial identities:
$
\displaystyle\sum_{k=0}^n \binom{n}{k} = 2^n
$
$
\displaystyle\sum_{k=0}^n k \binom{n}{k} = n 2^{n-1}
$

Calculate the sum:
$
\displaystyle\sum_{k=0}^{29} (k+1) \binom{29}{k} = \displaystyle\sum_{k=0}^{29} k \binom{29}{k} + \displaystyle\sum_{k=0}^{29} \binom{29}{k} = 29 \cdot 2^{28} + 2^{29} = (29 + 2) 2^{28} = 31 \cdot 2^{28}
$

Therefore, the original sum evaluates to:
$
30 \times 31 \times 2^{28} = 15 \times 31 \times 2^{29} = 465 \times 2^{29}
$

Hence, the value of $\alpha$ is $465$

Question 19: The sum of the series $2 \times 1 \times{ }^{20} \mathrm{C}_4-3 \times 2 \times{ }^{20} \mathrm{C}_5+4 \times 3 \times{ }^{20} \mathrm{C}_6-5 \times 4 \times$ ${ }^{20} \mathrm{C}_7+\ldots+18 \times 17 \times{ }^{20} \mathrm{C}_{20}$, is equal to

1) 34
2) ---
3) ---
4) ---

Correct answer: (1)

Solution:

$\begin{aligned}
(1+x)^{20}={ }^{20} C_0 & +{ }^{20} C_1 x+{ }^{20} C_2 x^2+{ }^{20} C_3 x^3 \\
& +{ }^{20} C_4 x^4+{ }^{20} C_5 x^5+\ldots .+{ }^{20} C_{20} x^{20} \\
\frac{(1+x)^{20}}{x^2}=\frac{1}{x^2}+ & \frac{20}{x}+{ }^{20} C_2+{ }^{20} C_3 x \\
& +{ }^{20} C_4 x^2+{ }^{20} C_5 x^3+\ldots .+{ }^{20} C_{20} x^{18}
\end{aligned}$

Differentiate w.r.t  $x$

$\begin{aligned}
& \frac{x^2 \cdot 20(1+x)^{19}+(1+x)^{20} \cdot 2 x}{x^4}=\frac{-2}{x^3}-\frac{20}{x^2}+{ }^{20} C_3 \\
& +2 \cdot{ }^{20} C_4 x+3 \cdot{ }^{20} C_5 x^2+\ldots+18^{20} C_{20} x^{17} \\
& \frac{2 x(1+x)^{19}(11 x+1)}{x^4}=\frac{-2}{x^3}-\frac{20}{x^2}+{ }^{20} C_3 \\
& +2 \cdot{ }^{20} C_4 x+3 \cdot{ }^{20} C_5 x^2+\ldots+18{ }^{20} C_{20} x^{17}
\end{aligned}$

Differentiate w.r.t $x$

$\begin{aligned}
& \frac{2\left[x^3(1+x)^{19} \cdot 11+x^3(11 x+1) \cdot 19(1+x)^{18}-(1+x)^{19}(11 x+1) \cdot 3 x^2\right]}{x^6} \\
& =\frac{6}{x^4}+\frac{40}{x^3}+2 \cdot 1 \cdot{ }^{20} C_4+3 \cdot 2^{20} C_5 x+\ldots+18 \cdot 17{ }^{20} C_{20} x^{16}
\end{aligned}$

Put $x=-1$ in the above equation

$\Rightarrow 0=6-40+R \Rightarrow R=34$
Hence, the answer is ($34$).

Question 20: If $\sum_{\mathrm{r}=0}^{10}\left(\frac{10^{\mathrm{r}+1}-1}{10^{\mathrm{r}}}\right) \cdot{ }^{11} \mathrm{C}_{\mathrm{r}+1}=\frac{\alpha^{11}-11^{11}}{10^{10}}$, then $\alpha$ is equal to:

1) 15
2) 11
3) 24
4) 20

Correct answer: (4)

Solution:

Given:
$\sum_{r=0}^{10} \left(\frac{10^{r+1} - 1}{10^r}\right) \cdot {}^{11}\mathrm{C}_{r+1} = \frac{\alpha^{11} - 11^{11}}{10^{10}}$

Simplify the term inside the sum:
$\frac{10^{r+1} - 1}{10^r} = \frac{10 \cdot 10^r - 1}{10^r} = 10 - \frac{1}{10^r}$

Rewrite the sum:
$\sum_{r=0}^{10} \left(10 - \frac{1}{10^r}\right) \cdot \binom{11}{r+1} = \sum_{r=0}^{10} 10 \cdot \binom{11}{r+1} - \sum_{r=0}^{10} \frac{1}{10^r} \cdot \binom{11}{r+1}$

Changing the index $k = r+1$, $k$ goes from 1 to 11:
\[
\sum_{r=0}^{10} \binom{11}{r+1} = \sum_{k=1}^{11} \binom{11}{k} = 2^{11} - 1 = 2047
\]

Substitute:
$= 10 \times 2047 - \sum_{r=0}^{10} \frac{1}{10^r} \cdot {}^{11}\mathrm{C}_{r+1}$

Use the binomial expansion:
$(1 + x)^{11} = \sum_{k=0}^{11} {}^{11}\mathrm{C}_k \cdot x^k$

Putting $x = \frac{1}{10}$:
$\left(1 + \frac{1}{10}\right)^{11} = \sum_{k=0}^{11} {}^{11}\mathrm{C}_k \cdot 10^{-k}$

Rewrite the sum inside subtraction:
\[
\sum_{r=0}^{10} \frac{1}{10^r} \cdot {}^{11}C_{r+1} 
= \sum_{k=1}^{11} {}^{11}C_k \cdot 10^{-(k-1)} 
= 10 \sum_{k=1}^{11} {}^{11}C_k \cdot 10^{-k}
\]

This equals:
$10 \left( \sum_{k=0}^{11} {}^{11}\mathrm{C}_k \cdot 10^{-k} - 1 \right) = 10 \left( \left(1 + \frac{1}{10}\right)^{11} - 1 \right)$

Simplify:
$10 \left( \left(\frac{11}{10}\right)^{11} - 1 \right)$

Substitute back:
$\sum_{r=0}^{10} \left(\frac{10^{r+1} - 1}{10^r}\right) \cdot {}^{11}\mathrm{C}_{r+1} = 10 \times 2047 - 10 \left(\left(\frac{11}{10}\right)^{11} - 1\right) = 10 \times 2047 - 10 \left(\frac{11}{10}\right)^{11} + 10$
$= 10 \times 2048 - 10 \left(\frac{11}{10}\right)^{11}$

Recall the original equation:
$\frac{\alpha^{11} - 11^{11}}{10^{10}} = 10 \times 2048 - 10 \left(\frac{11}{10}\right)^{11}$

Multiply both sides by $10^{10}$:
$\alpha^{11} - 11^{11} = 10^{11} \times 2048 - 10^{11} \times \left(\frac{11}{10}\right)^{11}$

Simplify the right side:
$= 2048 \times 10^{11} - 11^{11}$

Rearranged:
$\alpha^{11} = 2048 \times 10^{11}$

Since $2048 = 2^{11}$,
$\alpha^{11} = 2^{11} \times 10^{11} = (2 \times 10)^{11} = 20^{11}$

Taking the 11th root:
$\alpha = 20$

Hence, the correct answer is option (4).

Question 21: The product of the last two digits of $(1919)^{1919}$ is _________

1) 63
2) -
3) -
4) -

Correct answer: (1)

Solution:

We are asked to find the product of the last two digits of $(1919)^{19}$.  
To do this efficiently, we apply the Binomial Theorem by expressing the base in a convenient form close to a multiple of 10.

We write: $
1919 = 1920 - 1
$

So: $
(1919)^{19} = (1920 - 1)^{19}
$

By Binomial Theorem:

$
(1920 - 1)^{19} = \sum_{k=0}^{19} \binom{19}{k} \cdot (1920)^{19 - k} \cdot (-1)^k
$

We only need the last two digits, so we focus on the terms that contribute to lowest powers of 1920, since higher powers of 1920 are multiples of 100 and do not affect the last two digits.

So we retain only the last two terms of the expansion:

$
\begin{aligned}
T_{18} &= \binom{19}{18} \cdot (1920)^1 \cdot (-1)^{18} = 19 \cdot 1920 \cdot 1 = 36480 \\
T_{19} &= \binom{19}{19} \cdot (1920)^0 \cdot (-1)^{19} = 1 \cdot 1 \cdot (-1) = -1
\end{aligned}
$

Adding these two: $
36480 - 1 = 36479
$

The last two digits of $36479$ are: $
79
$

Product of Last Two Digits

Compute: $
7 \times 9 = 63
$

Hence, the answer is $63$.

Question 22: If $\sum_{k=0}^{n+1} \sum_{j=1}^{k+1} \sum_{i=1}^k{ }^i c_{j-1}($ where $n \in N)$ is $2^{n+3}-pn-q$, then $q-p=$

1) 4
2) -
3) -
4) -

Correct answer: (1)

Solution:

$\begin{aligned} & \sum_{k=0}^{n+1} \sum_{j=1}^{k+1}\left(\sum_{i=1}^k{ }^{i+1} C_j-{ }^i C_j\right)=\sum_{k=0}^{n+1} \sum_{j=1}^{k+1}\left({ }^{k+1} C_j-{ }^1 C_j\right)=\sum_{k=0}^{n+1}\left(2^{k+1}-2\right) \\ & =\sum_{k=0}^{n+1} 2^{k+1}-2 \sum_{k=0}^{n+1} 1=2 \cdot\left(2^{n+2}-1\right)-2(n+2)=2^{n+3}-2 n-6 .\end{aligned}$

Hence, the answer is 4.

Question 23: If $\sum_{\mathrm{r}=0}^5 \frac{{ }^{11} \mathrm{C}_{2 \mathrm{r}+1}}{2 \mathrm{r}+2}=\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}-\mathrm{n}$ is equal to______________.

1) 2035
2) -
3) -
4) -

Correct answer: (1)

Solution:

We are given
$\sum_{r=0}^5 \dfrac{{}^{11}C_{2r+1}}{2r+2} = \dfrac{m}{n}$, where $\gcd(m,n)=1$.

Consider the binomial expansion
$(1+x)^{11} = C_0 + C_1 x + C_2 x^2 + \cdots + C_{11} x^{11}$.

Integrate from $0$ to $1$:
$\int_0^1 (1+x)^{11} dx = \left[C_0 x + \dfrac{C_1 x^2}{2} + \dfrac{C_2 x^3}{3} + \cdots + \dfrac{C_{11} x^{12}}{12}\right]_0^1$

So, $\int_0^1 (1+x)^{11} dx = C_0 + \dfrac{C_1}{2} + \dfrac{C_2}{3} + \dfrac{C_3}{4} + \cdots + \dfrac{C_{11}}{12}$

But $\int_0^1 (1+x)^{11} dx = \dfrac{(2)^{12}-1}{12} = \dfrac{4096-1}{12} = \dfrac{4095}{12}$.

Now integrate from $-1$ to $0$:
$\int_{-1}^0 (1+x)^{11} dx = \left[C_0 x + \dfrac{C_1 x^2}{2} + \dfrac{C_2 x^3}{3} + \cdots \right]_{-1}^0$

This gives: $C_0 - \dfrac{C_1}{2} + \dfrac{C_2}{3} - \dfrac{C_3}{4} + \cdots - \dfrac{C_{11}}{12}$

Also, $\int_{-1}^0 (1+x)^{11} dx = \dfrac{1}{12}$.

Subtracting the two results:
$2\left(\dfrac{C_1}{2} + \dfrac{C_3}{4} + \dfrac{C_5}{6} + \dfrac{C_7}{8} + \dfrac{C_9}{10} + \dfrac{C_{11}}{12}\right)
= \dfrac{4095}{12} - \dfrac{1}{12} = \dfrac{4094}{12}$

Hence, $\dfrac{C_1}{2} + \dfrac{C_3}{4} + \dfrac{C_5}{6} + \dfrac{C_7}{8} + \dfrac{C_9}{10} + \dfrac{C_{11}}{12}
= \dfrac{4094}{24} = \dfrac{2047}{12}$

So, $m = 2047, n = 12$

Therefore, $m - n = 2047 - 12 = 2035$.

Hence, the answer is 2035.

Question 24: If the coefficients of $x^4, x^5$ and $x^6$ in the expansion of $(1+x)^n$ are in the arithmetic progression, then the maximum value of $n$ is :

1) 14
2) 21
3) 28
4) 7

Correct answer: (1)

Solution:

$\begin{aligned} & \text { Coeff. of } x^4={ }^n C_4 \\ & \text { Coeff. of } x^5={ }^n C_5 \\ & \text { Coeff. of } x^6={ }^n C_6 \\ & { }^n C_4,{ }^n C_5,{ }^n C_6 \ldots . A P \\ & 2 .{ }^n C_5={ }^n C_4+{ }^n C_6 \\ & 2=\frac{{ }^n C_4}{{ }^n C_5}+\frac{{ }^n C_6}{{ }^n C_5} \quad\left\{\frac{{ }^n C_r}{{ }^n C_r-1}=\frac{n-r+1}{r}\right\} \\ & 2=\frac{5}{n-4}+\frac{n-5}{6} \\ & 12(n-4)=30+n^2-9 n+20 \\ & n^2-21 n+98=0 \\ & (n-14)(n-7)=0 \\ & n_{\max }=14 \quad n_{\min }=7\end{aligned}$

Hence, the answer is option (1).

Question 25: If the constant term in the expansion of $\mathrm{\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}}, x \neq 0$, is $\mathrm{\alpha \times 2^8 \times \sqrt[5]{3}}$, then $25 \alpha$ is equal to :

1) 639
2) 724
3) 693
4) 742

Correct answer: (3)

Solution:

$\begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}}\left(\frac{3^{1 / 5}}{\mathrm{x}}\right)^{12-\mathrm{r}}\left(\frac{2 \mathrm{x}}{5^{1 / 3}}\right)^{\mathrm{r}} \\ & \mathrm{T}_{\mathrm{r}+1}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}}(3)^{\frac{12-\mathrm{r}}{5}}(2)^{\mathrm{r}}(\mathrm{x})^{2 \mathrm{r}-12}}{(5)^{\mathrm{r} / 3}} \\ & \mathrm{r}=6 \\ & \mathrm{~T}_7=\frac{{ }^{12} \mathrm{C}_6(3)^{6 / 5}(2)^6}{5^2}=\left(\frac{9 \times 11 \times 7}{25}\right) 2^8 \cdot 3^{1 / 5} \\ & 25 \alpha=693\end{aligned}$

Hence, the answer is the option (3).

Topic-Wise Weightage and Number of Questions

The table below shows the distribution of questions asked from the chapter Binomial Theorem in the January and April JEE Main 2026 exams. The table will help students to have an idea about the topics important from the point of view of the exam and on which they should focus․

Last 10 Years' Weightage of Binomial Theorem in JEE Main

The following table shows the topic-wise distribution of the number of questions from the topic of the binomial theorem asked per year over the last 10 years of the JEE Main exam. Candidates can use this to identify high-weightage topics and plan their preparation strategy effectively

Also Read: Last 10 Years JEE Main Question Paper

Important Formulas of Binomial Theorem for JEE Main 2027

1. Binomial Expansion:  $(a+b)^n=\binom{n}{0} a^n+\binom{n}{1} a^{n-1} b+\binom{n}{2} a^{n-2} b^2+\cdots+b^n$

2. General Term: The $(r+1)^{\text {th }}$ term in the expansion of $(a+b)^n$ is: $T_{r+1}=\binom{n}{r} a^{n-r} b^r$

3. Binomial Coefficient: $\binom{n}{r}=\frac{n!}{r!(n-r)!}$

4. Middle Term: 

• If $n$ is even, there is only one middle term:

$\left(\frac{n}{2}+1\right)^{\text {th }} \text { term }$

• If $n$ is odd, there are two middle terms:

$\left(\frac{n+1}{2}\right)^{t h} \text { and }\left(\frac{n+3}{2}\right)^{t h} \text { terms }$

5. Expansion of $(a-b)^n$: $(a-b)^n=\sum_{r=0}^n(-1)^r\binom{n}{r} a^{n-r} b^r$

6. Sum of Binomial Coefficients: $\sum_{r=0}^n\binom{n}{r}=2^n$

7. Alternating Sum of Binomial Coefficients: $\sum_{r=0}^n(-1)^r\binom{n}{r}=0$