Question

Find the sum of $2+4+6+....+200$.

Answer 1

Calculate the sum of the given series

Given series: $2+4+6+....+200$.

Thus, $2+4+6+....+200=2\left(1+2+3+....+100\right)$

It is clear that $\left(1+2+3+....+100\right)$ is in arithmetic progression.

The number of terms is, $n=100$.

The first term, $a=1$.

The last term, $l=100$.

The sum of $n$ terms with the first term $a$ and the last term $l$ can be given by, $\mathit{S}\mathbf{=}\frac{\mathbf{n}\left(a+l\right)}{\mathbf{2}}$.

So, $\left(1+2+3+....+100\right)=\frac{n}{2}\left(a+l\right)$

$$\begin{gathered} \Rightarrow \left(1+2+3+....+100\right)=\frac{100}{2}\left(1+100\right) \\ \Rightarrow 2\left(1+2+3+....+100\right)=2\times \frac{100}{2}\times 101 \\ \Rightarrow 2+4+6+....+200=100\times 101 \\ \Rightarrow 2+4+6+....+200=10100 \end{gathered}$$

Therefore, the sum of the given series is $10100$.