Question

For given series $1^2+2\times 2^2+3^2+2\times 4^2+5^2+2\times 6^2+\dots ,\text{ if }{S}_n$ 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 $P(n):S_n=\{\begin{array}{l}\frac{n(n+1{)}^2}{2},\text{ when }n\text{ is even }\\ \frac{n^2(n+1)}{2},\text{ when }n\text{ is odd }\end{array}$
Also, note that any term $T_n$ of the series is given by
$T_n=\{\begin{array}{l}{n}^2,\text{ if }n\text{ is odd }\\ 2n^2,\text{ if }n\text{ is even }\end{array}$
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,
$P(k+1):S_{k+1}=1^2+2\times 2^2+\dots +k^2+2\times (k+1{)}^2$
$-\frac{k^2(k+1)}{2}+2\times (k+1{)}^2$
$\left[\text{ as }k\text{ is odd },1^2+2\times 2^2+\cdots +k^2=k^2\frac{(k+1)}{2}\right]$
$=\frac{(k+1)}{2}\left[k^2+4(k+1)\right]$
$=\frac{k+1}{2}\left[k^2+4k+4\right]$
$=\frac{k+1}{2}(k+2{)}^2$
$=(k+1)\frac{[(k+1)+1{]}^2}{2}$
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, $P(k+1):S_{k+1}=1^2+2\times 2^2+\dots +2\cdot k^2+(k+1{)}^2$
$=\frac{k(k+1{)}^2}{2}+(k+1{)}^2$
$[$ as $k$ is even, $1^2+2\times 2^2+\dots +2k^2=k\frac{(k+1{)}^2}{2}]$
$=\frac{(k+1{)}^2(k+2)}{2}=\frac{(k+1{)}^2((k+1)+1)}{2}$
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 $n$.