Question

The sum of first n terms of the given 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 will be

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is B

When n is odd, the last term i.e., the $n^{th}$ term will be $n^2$

in this case n-1 is even and so the sum of the first n-1

term of the series is obtained by replacing n by n-1 in the

given formula and so is $\frac{1}{2}(n-1)n^2$.

Hence the sum of the n terms

= (the sum of n-1 terms) + the $n^{th}$ term

= $\frac{1}{2}(n-1)n^2$ + $n^2$ = $\frac{1}{2}(n+1)n^2$

Trick : check for n=1,3. Here $S_1$ = 1, $S_3$ = 18 which gives (b).