Answer
$\sum_{k=1}^{5}(-1)^{k-1}\frac{k+1}{(k+2)(k+3)}$
Work Step by Step
1. We can represent the alternating signs using $(-1)^{k-1}$
2. We can express the original summation as:
$\sum_{k=1}^{5}(-1)^{k-1}\frac{k+1}{(k+2)(k+3)}$
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