question_answer

The sum of the first\[n\]terms of the series \[{{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+....\]is \[\frac{n{{(n+1)}^{2}}}{2},\] when n is even. When n is odd, the sum is

A) \[\frac{3n(n+1)}{2}\]     

B)        \[\frac{{{n}^{2}}(n+1)}{2}\]        

C)        \[\frac{n{{(n+1)}^{2}}}{4}\]        

D)        \[{{\left[ \frac{n(n+1)}{4} \right]}^{2}}\]

Correct Answer: B

Solution :

Letthen                                                                                 Puttingwe get the desired sum is Alternate Solution Given that the sum of n terms of given series is if n is even. Letbe odd i.e.,. Then, term term