Question

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

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

Answer 1

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

$S=1^2-2^2+3^2-4^2......{(-1)}^{n-1}{n}^2$

$=1^2-2^2+3^2....n^2-2(2^2+4^2....{(n-1)}^2)$ (Odd)

OR

$=1^2-2^2+3^2....{(n-1)}^2-2(2^2+4^2....{(n-2)}^2)-n^2$ (Even)

$=\frac{n(n+1)(2n+1)}{6}-8(1^2+2^2....{\left(\frac{n-1}{2}\right)}^2)$ (Odd)

OR

$\frac{(n-1)\left(n\right)(2n-1)}{6}-8(1^2+2^2.....{(\frac{n}{2}-1)}^2)-n^2$ (Even)

$=\frac{n(n+1)(2n+1)}{6}-4\frac{\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)\left(n\right)}{6}$ (Odd)

OR

$\frac{n(n-1)(2n-1)}{6}-4\frac{\left(\frac{n-2}{2}\right)\left(\frac{n}{2}\right)(n-1)}{6}-n^2$ (Even)

$=\frac{n(n+1)}{6}[2n+1-2(n-1)]$ (Odd)

OR

$\frac{n(n-1)}{6}[2n-1-2(n-2)]-n^2$ (Even)

$=\frac{n(n+1)}{6}\times 3$ (Odd)

OR

$\frac{n(n-1)}{6}\times \left(3\right)-n^2$

$=\frac{n(n+1)}{2}$ (Odd)

OR

$\frac{n(n-1)}{2}-n^2$ (Even)

$\Rightarrow -\frac{n(n+1)}{2}$ (Even)

Option 1 is correct if n is even while,

Option 2 is correct if n is odd.