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Question

The sum of first n terms of the series 12+2.22+32+2.42+52+5.62+... is n(n+1)22 where n is even. When n is odd the sum is

A
n2(n+1)2
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B
n(n+1)22
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C
(n(n+1)2)2
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D
n(n+1)2
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Solution

The correct option is C n2(n+1)2
When n is odd, last term will be n2,
Then the sum is
12+2.22+32+2.42+52+5.62+...+2(n1)2+n2
=n(n+1)22+n2
[ Replacing n by n1 in n(n+1)22]
=n3n2+2n22=n3+n22=n2(n+1)2

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