Question

The sum to n terms of the series, where n is an even number: $1^2-2^2+3^2-4^2+5^2-6^2+\cdots$:

• A. $n(n+1)$
• B. $\frac{n(n+1)}{2}$
• C. $-\frac{n(n+1)}{2}$
• D. None of these

Answer 1

The correct option is C $-\frac{n(n+1)}{2}$

Solve by putting different values for n.
When n = 2, given expression = -3
and when n = 4, given expression = -10
Only option (c) is correct for both cases, hence is a corrct answer.

Alternate (Conventional) Approach:
$1^2-2^2+3^3-4^2+5^2-6^2+7^2-8^2+\cdots$
$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+(7+8)(7-8)+\cdots$
$=-(1+2)-(3+4)-(5+6)\cdots$
$=-\left[\right(1+2)+(3+4)+(5+6)+\cdots ]$
$=-[1+2+3+4+5+6+\cdots ]=-\left[\frac{n(n+1)}{2}\right]$