Sum of Common Series - Practice Questions & MCQ

The sum of Common Series
1. The sum of the first n-natural numbers

$
1+2+3+4+5+\ldots \ldots .+n=\frac{n(n+1)}{2}
$
Use the formula, sum of $n-$ term of an AP $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+1) ;$ where, $\mathrm{a}=1, \mathrm{l}=\mathrm{n}$ and number of term is n

$
\begin{aligned}
& \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(1+\mathrm{n}) \\
& \Rightarrow \sum_{\mathrm{n}=1}^{\mathrm{n}} \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}
\end{aligned}
$

2. The sum of the first n-odd natural number

$
\begin{aligned}
& 1+3+5+7+\ldots \ldots \text { upto } \mathrm{n} \text { term }=\frac{\mathrm{n}}{2}(2 \cdot 1+(\mathrm{n}-1) \cdot 2) \\
& \Rightarrow \sum(2 \mathrm{n}-1)=\mathrm{n}^2
\end{aligned}
$
 

The sum of Common Series
1. Sum of first n-even natural number

$
2+4+6+8+\ldots \ldots \ldots=n / 2[2 \times 2+(n-1) 2]=n(n+1)
$

2. Sum of the squares of first n-natural numbers

$
1^2+2^2+3^2+4^2+\ldots \ldots \ldots+n^2=\frac{n(n+1)(2 n+1)}{6}
$
Proof:
we have, $n^3-(n-1)^3=3 n^2-3 n+1$; by changing $n$ to $(n-1)$, we get

$
\begin{aligned}
& (n-1)^3-(n-2)^3=3(n-1)^2-3(n-1)+1 \\
& (n-2)^3-(n-3)^3=3(n-2)^2-3(n-2)+1 \\
& \\
& 3^3-2^3=3 \times 3^2-3 \times 3+1 \\
& 2^3-1^3=3 \times 2^2-3 \times 2+1 \\
& 1^3-0^3=3 \times 1^2-3 \times 1+1
\end{aligned}
$
Hence, in addition,

$
\begin{aligned}
n^3 & =3\left(1^2+2^2+3^2+\ldots \ldots+\mathrm{n}^2\right)-3(1+2+3+\ldots \ldots+\mathrm{n})+\mathrm{n} \\
& =3 \mathrm{~S}-\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}+\mathrm{n} \\
3 \mathrm{~S} & =n^3-n+\frac{3 n(n+1)}{2} \\
& =n(n+1)\left(n-1+\frac{3}{2}\right) \\
\Rightarrow \mathrm{S} & =\frac{n(n+1)(2 n+1)}{6}
\end{aligned}
$
 

Sum of Common Series
1. Sum of the cube of first $n$-natural numbers

$
1^3+2^3+3^3+4^3+\ldots \ldots \ldots+\mathrm{n}^3=\left\{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right\}^2
$
Proof:
We have,

$
\begin{aligned}
& n^4-(n-1)^4=4 n^3-6 n^2+4 n-1 \\
& (n-1)^4-(n-2)^4=4(n-1)^3-6(n-1)^2+4(n-1)-1 \\
& (n-2)^4-(n-3)^4=4(n-2)^3-6(n-2)^2+4(n-2)-1 \\
& \vdots \\
& 3^4-2^4=4 \times 3^3-6 \times 3^2+4 \times 3-1 \\
& 2^4-1^4=4 \times 2^3-6 \times 2^2+4 \times 2-1 \\
& 1^4-0^4=4 \times 1^3-6 \times 1^2+4 \times 1-1
\end{aligned}
$
Hence, by addition

$
\begin{aligned}
n^4 & =4 S-6\left(1^2+2^2+\cdots+n^2\right)+4(1+2+\cdots+n)-n \\
4 S & =n^4+n+6\left(1^2+2^2+\cdots+n^2\right)-4(1+2+\cdots+n) \\
& =n^4+n+n(n+1)(2 n+1)-2 n(n+1) \\
& =n(n+1)\left(n^2-n+1+2 n+1-2\right) \\
& =n(n+1)\left(n^2+n\right) \\
S & =\frac{n^2(n+1)^2}{4}=\left\{\frac{n(n+1)}{2}\right\}^2
\end{aligned}
$