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Sum of series 4.


∞∑n=2(−1)nn2+n−2

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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