An algebra problem by Naren Bhandari

Consider:

$n^2 - (n-1)^2 \equiv (n - (n-1))(n + (n-1)) \equiv 1×(n + (n-1)) \equiv$

$\equiv n + (n-1)$

Let's apply the identity above for n = 100, 98, 96, ..., 4, 2

We get:

$100^2 - 99^2 + 98^2 - 97^2 + ... + 4^2 - 3^2 + 2^2 - 1^2 =$

$= 100 + 99 + 98 + 97 + ... + 4 + 3 + 2 + 1 = \frac { 100 × (100 + 1) } {2} = \boxed {5050}$

$\begin{aligned} S & = 100^2 - 99^2 + 98^2 - 97^2 + 96^2 - 95^2 + \cdots + 2^2-1^2 \\ & = (100-99)(100+99) + (98-97)(98+97) + (96-95)(96+95) + \cdots + (2-1)(2+1) \\ & = 199 + 195 + 191 + \cdots + 3 \\ & = \underbrace{3}_{k=1} + \underbrace{7}_{2} + \underbrace{11}_{3} + \cdots + \underbrace{199}_{n=50} & \small \color{#3D99F6} \text{General term: } 4k -1 \\ & = \frac {50(3+199)}2 & \small \color{#3D99F6} \text{Sum of AP: } S=\frac {n(a+l)}2 \\ &=\boxed{5050} \end{aligned}$

This problem seems like one of my problem. You can check it here