Question

The sum of $n$ terms of the series $1^2-2^2+3^2-4^2+5^2-6^2+...$ is

• A. $\frac{-n(n+1)}{2}$ if $n$ is even
• B. $\frac{n(n+1)}{2}$ if $n$ is odd
• C. $-n(n+1)$ if $n$ is even
• D. $\frac{n(n+1)(2n+1)}{6}$ if $n$ is odd

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{-n(n+1)}{2}$ if $n$ is even
B $\frac{n(n+1)}{2}$ if $n$ is odd

If n is even

$1^2+2^2+3^2.....n^2-2(2^2+4^2...n^2)$ = $n(n+1)(2n+1)/6-8\left(n/2\right(n/2+1\left)\right(n+1\left)\right)/6$

=$-n(n+1)/2$

if nis odd

$1^2+2^2+3^2.....n^2-$2($2^2+4^2...(n-1{)}^2$)=$n(n+1)(2n+1)/6-8\left(\right(n-1)/2\times (n+1)/2\times (n\left)\right)/6$$-2n^2$

$=n(n+1)/2$