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Some Standard Expansions - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
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Some Standard Expansions (Part 2) is considered one of the most asked concept.
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78 Questions around this concept.
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Remainder when $7^{100}$ is divided by 25 is
If $(\sqrt{2}+1)^6=I+F_{\text {where }} \leq F<1$ and $I \in N{\text { then the value of } \mathrm{I}}$ is
If $(1+x)^n-C_0+C_1 x+C_2 x^2+C_3 x^3+\cdots+C_n x^n$, then $\mathrm{C}_0 \mathrm{C}_1+\mathrm{C}_1 \mathrm{C}_2+\cdots+\mathrm{C}_{\mathrm{n}-1} \mathrm{C}_{\mathrm{n}}$ is equal to
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If $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots \ldots .+C_n x^n$, then the value of $2 \mathrm{C}_0+4 \mathrm{C}_1+6 \mathrm{C}_2+\ldots \ldots \ldots+2(\mathrm{n}+1) \mathrm{C}_{\mathrm{n}}$ will be
If $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n$, then the value of $\sum_{\mathrm{k}=0}^{\mathrm{n}}(\mathrm{k}+1)^2 \cdot C_k$
If $\{x\}$ denotes the fractional part of $x$, then $\left\{\frac{3^{2 n}}{8}\right\}, n \in N$ is
If $\left(1+2 x+3 x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$, then $a_1$ equals
If $\left(1+x-2 x^2\right)^5=1+a_1 x+a_2 x^2+\ldots+a_{10} x^{10}$, then $a_2+a_4+a_6+\ldots+a_{10}$
If $\mathrm{a}_{\mathrm{k}}$ is the coefficient of $\mathrm{x}^{\mathrm{k}}$ in the expansion of $\left(1+\mathrm{x}+\mathrm{x}^2\right)^{\mathrm{n}}$ for $\mathrm{k}=0,1,2, \ldots \ldots, 2 \mathrm{n}$ then $5 \cdot \mathrm{a}_1+10 \cdot \mathrm{a}_2+15 \cdot \mathrm{a}_3+\ldots \ldots+10 \cdot \mathrm{na}_{2 \mathrm{n}}$
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Try NowThe number $101^{100}-1$ is divisible by
Concepts Covered - 2
Some Standard Expansions (Part 1)
We know the binomial expansion,
$
(\mathrm{x}+\mathrm{y})^{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_0 \mathrm{x}^{\mathrm{n}}+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}^{\mathrm{n}-1} \mathrm{y}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^{\mathrm{n}-2} \mathrm{y}^2+\cdots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{y}^{\mathrm{n}}
$
i.e. $\quad(x+y)^n=\sum_{r=0}^n{ }^n C_r x^{n-r} y^r$
1. Replace ' $y$ ' with ' $-y$ ' in the binomial expansion, we get
$
\begin{aligned}
& (x-y)^n={ }^n C_0 x^n-{ }^n C_1 x^{n-1} y+{ }^n C_2 x^{\mathrm{n}-2} y^2-\cdots+(-1)^{\mathrm{r} n} C_r x^{\mathrm{n}-\mathrm{r}} y^{\mathrm{r}}+\cdots+(-1)^{\mathrm{n}{ }^n} C_n y^{\mathrm{n}} \\
& \text { or } \quad(\mathrm{x}-\mathrm{y})^{\mathrm{n}}=\sum_{\mathrm{r}=0}^{\mathrm{n}}(-1)^{\mathrm{r} n} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{n}-\mathrm{r}} y^{\mathrm{r}}
\end{aligned}
$
2. In the binomial expansion, $(x+y)^n$ replace ' $x$ ' by 1 and ' $y$ ' by $x$
$
\begin{aligned}
& (1+\mathrm{x})^{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_0 \mathrm{x}^0+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}^1+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+\cdots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}+\cdots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \\
& \text { or } \quad(1+\mathrm{x})^{\mathrm{n}}=\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}
\end{aligned}
$
3. In the binomial expansion, $(x+y)^n$ replace ' $x$ ' by ' 1 ' and ' $y$ ' by ' $-x$ '
$
\begin{aligned}
& (1-\mathrm{x})^{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_0 \mathrm{x}^0-{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}^1+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2-\cdots+(-1)^{\mathrm{r}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}+\cdots+(-1)^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \\
& \text { or } \quad(1-x)^n=\sum_{r=0}^n(-1)^r{ }^n C_r x^r
\end{aligned}
$
Some Standard Expansions (Part 2)
4. Addition: $(x+y)^n+(x-y)^n$
$
(x+y)^n+(x-y)^n=2\left[{ }^n C_0 x^n y^0+{ }^n C_2 x^{n-2} y^2+{ }^n C_4 x^{n-4} y^4+\ldots .\right]
$
If ' $n$ ' is odd then number of terms is $\frac{n+1}{2}$
If $n^{\prime}$ is even then number of terms is $\frac{n^2}{2}+1$
5. Subtraction: $(x+y)^n-(x-y)^n$
$
(x+y)^n-(x-y)^n=2\left[{ }^n C_1 x^{n-1} y^1+{ }^n C_3 x^{n-3} y^3+{ }^n C_5 x^{n-5} y^5+\ldots \ldots\right]
$
If $n$ is odd, then the number of terms is $\frac{n+1}{2}$
If $n$ is even, then the number of terms is $\frac{n}{2}$
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