Answer
$\sum_{k=1}^{20}\left((-1)^{k-1}(\frac{1}{k^2})\right)$
Work Step by Step
If we rewrite the first term as $\frac{1}{1}$, we can say that the absolute value of the k th term in the series is $\frac{1}{k^2}$.
The 1st term is $\frac{1}{1^2}$
The 2nd term is $\frac{1}{2^2}$
The last term is $\frac{1}{20^2}$, so here, k=20.
k is changing from 1 to 20 with a step of 1.
If we look at the k th term, where k is an odd number, the fraction will be with a positive sign, and if k is even, the fraction will be with a negative sign. This means, that we have to multiply each fraction by $(-1)^{k-1}$. If k is odd, this number will be simply 1, which means there will be a positive sign in front of the fraction. If k is even this number will be -1, which means there will be a negative sign in front of the fraction.
Therefore the sum can written like:
$\sum_{k=1}^{20}\left((-1)^{k-1}(\frac{1}{k^2})\right)$
