Q. 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 $\frac{n(n + 1)^{2}}{2}$ where $n$ 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) $\left\lbrack \frac{n(n + 1)}{2} \right\rbrack^{2}$
Explanation
(b)
The some of n terms of given series=$\frac{n(n + 1)^{2}}{2}$ if $n$ is even. Let $n$ is odd $ie$, $n = 2m + 1$
Then, $S_{2m + 1} = S_{2m} + (2m + 1)th$ term
$= \frac{(n - 1)n^{2}}{2} + nth$ term
$= \frac{(n - 1)n^{2}}{2} + n^{2}\ \ \ \ \ \lbrack\because n\ is\ odd = 2m + 1\rbrack$
$= n^{2}\left\lbrack \frac{n - 1 + 2}{2} \right\rbrack = \frac{(n + 1)n^{2}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.