SOLUTION
⇒∞∑n=2(−1)nn2+2n−n−2
⇒∞∑n=2(−1)nn(n+2)−(n+2)
⇒∞∑n=2(−1)n(n−1)(n+2)
⇒∞∑n=2[(−1)n−3(n−1)+(−1)n3(n+2)]
⇒13∞∑n=2[(−1)n+1n−1+(−1)nn+2]
⇒13∞∑n=2(−1)n+1n−1−13∞∑n=2(−1)n−1n+2⋯(1)
We have series of log(1+x) as
⇒log(1+x)=x−x22+x33−x44+⋯
Using this in (1) , we will get
⇒−13[log(2)]−13[log2−1+12−13]
⇒−23log2−16+19+13
⇒−23log2+−3+2+618
⇒−23log2+518
SOLUTION: ⇒−23log2+518
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