Fill in the blank so that the following equations are true: m Ση=0 i=1 100 Σ2 = 0 i=3

Given series is

$\sum _{i=1}^{n}n$

Expanding, we get

$\sum _{i=1}^{n}n=1+2+3+...+n$

The above series is arithmetic series with n number and its common difference is $d=1$

Since sum of arithmetic series is

$S_n=\frac{n}{2}\left(2a+(n-1)d\right)$

where $a$ is the first element.

We have $a=1$

So we have

$$\begin{gathered} S_n=\frac{n}{2}\left(2+(n-1)\right) \\ \Rightarrow S_n=\frac{n}{2}(n+1) \end{gathered}$$

Hence we can say

$\sum _{i=1}^{n}n=\frac{n(n+1)}{2}$