A simple problem

$\sum _{ m=1 }^{ 20 }{ \sin { \left( \cfrac { m\pi }{ 2 } \right) } }$

$=\sin { \left( \cfrac { \pi }{ 2 } \right) } +\sin { \left( \pi \right) } +\sin { \left( \cfrac { 3\pi }{ 2 } \right) } +\dots +\sin { \left( 10\pi \right) }$

As $\sin { \left( x \right) } =\sin { \left( x+2n\pi \right) } ~\forall ~n\in \mathbb{Z}$ :

$=5\left[ \sin { \left( \cfrac { \pi }{ 2 } \right) } +\sin { \left( \pi \right) } +\sin { \left( \cfrac { 3\pi }{ 2 } \right) } +\sin { \left( 2\pi \right) } \right]$

$=5\left[ 1+0-1+0 \right]$

$=\large\color{#20A900}{\boxed{0}}\\$

Because sin is cyclic, starting over after m=4 (because after 2*pi, sin repeats itself) in this particular equation, you can find the sum from m=1 to 4 and then multiply by 20/4 = 5. At m=1, the equation is sin(pi/2), which is 1. At m=2, the equation equals 0. At 3, it equals -1. Finally at 4, you get 0. The sum of these 4 is 0, so whole sum is 5x0 = 0.

The pattern of the sequence is 1, 0, -1, 0 and repeats every 4 terms, which means every 4 terms =0. For every multiple of 4 terms the sum would be 0 and if there were any remainders you would start the series from there.
Eg 26÷4 would have remainder 2, 0+(1+0) the sum would be 1

I dont understand plzz is there any simple way

Here from (1 to 20) there are 10 odd multiple of π/2 which are having values (1,-1,1,-1.......-1) these all gets cancelled.and there are 10 even multiple of π/2 which are having values (0,0,0,0....0) and hence adding all the values we get result as zero.

The positive cycle will always cancel the negative cycle for a full sinosoidal wave; thus 0