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Question

The sum of n terms of the series 12−22+32−42+52−62+... is

A
n(n+1)2 if n is even
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B
n(n+1)2 if n is odd
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C
n(n+1) if n is even
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D
n(n+1)(2n+1)6 if n is odd
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Solution

The correct options are
A n(n+1)2 if n is even
B n(n+1)2 if n is odd
If n is even
12+22+32.....n22(22+42...n2) = n(n+1)(2n+1)/68(n/2(n/2+1)(n+1))/6

=n(n+1)/2

if nis odd
12+22+32.....n22(22+42...(n1)2)=n(n+1)(2n+1)/68((n1)/2×(n+1)/2×(n))/62n2

=n(n+1)/2

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