Question

${100}^2-{99}^2+{98}^2-{97}^2+{96}^2+{95}^2+....=S$ find $S=?$

Answer 1

$S={100}^2-{99}^2+{98}^2-{97}^2+{96}^2-{95}^2+{94}^2+.....$

$=({100}^2-{99}^2)+({98}^2-{97}^2)+({96}^2-{95}^2)+....$

$=(100-99)(100+99)+(98-97)(98+97)+(96-95)(96+95)+....$

$=1(100+99)+1(98+97)+1(96+95)+....$

$=100+99+98+97+96+95+.....$

If $S$ is finite $S=100+99+98+97+96+....+1$ written in reverse order

$\Rightarrow S=1+2+3+....96+97+98+99+100$

Sum$=\frac{n(n+1)}{2}$,(sum of first n natural numbers)

here $n=100$

$S=\frac{100(100+1)}{2}=\frac{100\times 101}{2}=50\times 101=5050$