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Question: The sum of the first n terms of the series 1<sup>2</sup> + 2.2<sup>2</sup> + 3<sup>2</sup> + 2.4<sup...

The sum of the first n terms of the series 12 + 2.22 + 32 + 2.42 + 52 + 2.62 + . . . is, when n is even. When n is odd, the sum is

A

n2(n+1)2\frac { n ^ { 2 } ( n + 1 ) } { 2 }

B

n(n+1)(2n+1)6\frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 }

C

D

Answer

n2(n+1)2\frac { n ^ { 2 } ( n + 1 ) } { 2 }

Explanation

Solution

If n is odd, n – 1 is even. Sum of (n – 1) terms will be.

The nth term will be n2. Hence the required sum

= n2(n−1)2\frac { \mathrm { n } ^ { 2 } ( \mathrm { n } - 1 ) } { 2 } +n2 =.