Question

The sum to n terms of the series \[1^2-2^2+3^2-4^2+5^2-6^2........\]

Answer 1

My approach:

(1+(n-1)2)^2 = (2n-1)^2

(2+(n-1)2)^2 = (2n)^2


4n^2-4n+1 - 4n^2 = -4n+1

\[\sum_{1}^{n}-4k+1 \]

Answer 2

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

-2n^2 - 2n + n

= -2n^2 - n

Answer 3

is it finite series or infinite series

Answer 4

Hint: + and - terms differently and try it.