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Sum of n terms the series : 1^2-2^2+3^2-...

Sum of `n` terms the series : `1^2-2^2+3^2-4^2+5^2-6^2+`

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Given,
`1^2-2^2+3^2-4^2+5^2-6^2+`
If n is even
`1^2+2^2+3^2.....n^2−2(2^2+4^2...n^2) `
` = [n(n+1)(2n+1)]/6−8(n/2(n/2+1)(n+1))/6 `
`=(−n(n+1))/2`
...
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