Question

For given series $ 1^2+2 \times 2^2+3^2+2 \times 4^2+5^2+2 \times 6^2+\ldots, \text { if } S_n $ is the sum of $n$ terms, then

• A. $S_n=\frac{n(n+1)^2}{2}$, if $n$ is even
• B. $S_n=\frac{n^2(n+1)}{2}$, if $n$ is odd
• C. Both (a) and (b) are true
• D. Both (a) and (b) are false

Answer 1

Answer: (C)

Let $P(n): S_n=\begin{cases}\frac{n(n+1)^2}{2}, \text { when } n \text { is even } \\ \frac{n^2(n+1)}{2}, \text { when } n \text { is odd }\end{cases}$
Also, note that any term $T_n$ of the series is given by
$T_n=\begin{cases} n^2, \text { if } n \text { is odd } \\ 2 n^2, \text { if } n \text { is even } \end{cases}$
We observe that $P(1)$ is true, since
$P(1): S_1=1^2=1=\frac{1 \cdot 2}{2}=\frac{1^2 \cdot(1+1)}{2}$
Assume that $P(k)$ is true for some natural number $k$, i.e.,
Case I When $k$ is odd, then $k+1$ is even. We have,
$ P(k+1): S_{k+1}=1^2+2 \times 2^2+\ldots+k^2+2 \times(k+1)^2$
$-\frac{k^2(k+1)}{2}+2 \times(k+1)^2$
$ {\left[\text { as } k \text { is odd }, 1^2+2 \times 2^2+\cdots+k^2=k^2 \frac{(k+1)}{2}\right] }$
$=\frac{(k+1)}{2}\left[k^2+4(k+1)\right]$
$=\frac{k+1}{2}\left[k^2+4 k+4\right] $
$=\frac{k+1}{2}(k+2)^2$
$=(k+1) \frac{[(k+1)+1]^2}{2}$
So, $P(k+1)$ is true, whenever $P(k)$ is true, in the case when $k$ is odd.
Case II When $k$ is even, then $k+1$ is odd.
Now, $ P(k+1): S_{k+1} =1^2+2 \times 2^2+\ldots+2 \cdot k^2+(k+1)^2 $
$ =\frac{k(k+1)^2}{2}+(k+1)^2$
$\left[\right.$ as $k$ is even, $\left.1^2+2 \times 2^2+\ldots+2 k^2=k \frac{(k+1)^2}{2}\right]$
$=\frac{(k+1)^2(k+2)}{2}=\frac{(k+1)^2((k+1)+1)}{2}$
Therefore, $P(k+1)$ is true, whenever $P(k)$ is true for the case when $k$ is even.
Thus, $P(k+1)$ is true whenever $P(k)$ is true for any natural number $k$. Hence, $P(n)$ true for all natural numbers $n$.