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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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The correct Answer is:
`-(n(n+1))/2`
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Assertion: If n is odd then the sum of n terms of the series 1^2+2xx2^2+3^2+2xx4^2+5^2+2xx6^2+7^2+…is (n^2(n+1))/2. If n is even then the sum of n terms of the series. 1^2+2xx2^2+3^2+2xx4^2+5^2+2xx6^2+…..is (n(n+1)^2)/2 (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true R is not te correct explanation of A (C) A is true but R is false. (D) A is false but R is true.

Find the sum of n terms of the series 1^3+3.2^2+3^3+3.4^2+5^3+3.6^2+....... when (i)n is odd (ii)n is even

Write the sum of the series 1^2-2^2+3^2-4^2+5^2-6^2...........+(2n-1)^2-(2n)^2dot

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