Case: $\sum _{n=1}^{\infty}(-1)^{n+1}(2n-1)$ Question: $S_n =\ ?$ My partial solution and investigation:
$S_n = \sum _{k=1}^n (-1)^{n+1}(2n-1) = 1 -3+5-7+9-11+\cdots+(-1)^{n+1}(2n-1)$
I see there is a sum of two arithmetic sequences, but I don't know how to take advantage of this fact. After using the formula $S = S_a + S_b = \frac{(a_1+a_n)n}{2} + \frac{(b_1+b_n)n}{2}$ I got a wrong result $-n+(-1)^{n+1}n$.
The correct answer:
$(-1)^{n+1}n$