## 98 Powers of Two

Find the value of the following expression:
$\left (98^2-97^2+96^2-95^2+\cdots+2^2-1^2\right )^2$
Source: mathcontest.olemiss.edu 4/9/2012

SOLUTION
We simplify the expression by using the identity $a^2-b^2=\left (a+b\right )\left (a-b\right )$
$\left [\left (98+97\right )\left (98-97\right )+\left (96+95\right )\left (96-95\right )+\left (94+93\right )\left (94-93\right )+\cdots+\left (4+3\right )\left (4-3\right )+\left (2+1\right )\left (2-1\right )\right ]^2$
$\left [\left (195\right )\left (1\right )+\left (191\right )\left (1\right )+\left (187\right )\left (1\right )+\cdots+\left (7\right )\left (1\right )+\left (3\right )\left (1\right )\right ]^2$
$\left (195+191+187+\cdots+7+3\right )^2$
We calculate $S=195+191+187+\cdots+7+3$ as follows
$3=3$
$7=3+1\left (4\right )$
$11=3+2\left (4\right )$
$15=3+3\left (4\right )$
$\cdots$
$187=3+46\left (4\right )$
$191=3+47\left (4\right )$
$195=3+48\left (4\right )$
The sum $S$ is made up of 3’s and 4’s. How many 3’s are there?
$\left (195-3\right )\div 4+1=48+1$
$=49$
$S=49\left (3\right )+4\left (1+2+3+\cdots+46+47+48\right )$
$=147+4\left [\frac{\left (48+1\right )48}{2}\right ]$
$=147+4\left (49\right )24$
$=147+4704$
$=4851$
The value of the expression is
$4851^2=23532201$

Answer: $23532201$.