An algebra problem by Lakkoju Dilip

$100^2 - 99^2 + 98^2 - 97^2 -------------3^2 + 2^2 - 1^2$

$= [(100 + 99)(100 - 99)+(98 + 97)(98-97)+(96+95)(96-95)...(2+1)(2-1) ]$

$= [ 199 + 195 + 191........3)$

This arithmetic series with first term, $a =199$ , common difference $-4$ and last term $L = 3$ and number of terms, $n = 50$

sum of $50$ terms $= 50/2 [ a + L]$

$= 25(199 + 3)$

$= 25(202)$

$= 5050$

Gauss is said to have used a similar method to add the first $100$ positive integers as a small child, i.e. find the $100$ triangular number, which is $5050.$ In fact, as you are about to see, the result in this problem is the same.

$S = 100^2 - 99^2 + 98^2 - 97^2 + ... + 2^2 - 1^2 \\ \\ = (100+99)(100-99) + (98+97)(98-97) +\\ ... + (2+1)(2-1) \\ \\ = 199 + 195 + ... + 7 + 3 \\ \\ = (199 + 3) + (195 + 7) + ... + (103 + 99) \\ \\ = 202\times 25 \\ \\ = \boxed{5050}$

We know a^2-b^2=(a+b)(a-b) Hence, 100^2-99^2=(100+99)(100-99)=100+99

99^2-98^2=(99+98)(99-98)=99+98 ........ ........ 2^2-1^2=(2+1)(2-1)=2+1

To find the sum we just need to evaluate 100+99+98+.....+3+2+1

So the total sum is 100(100+1)/2=50*101=5050