Question

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. $\frac{n^2(n+1)}{2}$
• B. ${\left[\frac{n(n+1)}{2}\right]}^2$
• C. $\frac{3n(n+1)}{2}$
• D. $\frac{n(n+1{)}^2}{4}$

Answer 1

Answer: (A), (D)

$\frac{n(n+1{)}^2}{4}$
$\frac{n^2(n+1)}{2}$

For odd, let $n=2m+1$
$\Rightarrow S_{2m+1}=S_{2m}+{(2m+1)}^{th}term$ $...\left(1\right)$
Also, given that $S_{2m}=\frac{m{(m+1)}^2}{2}$.
Since, $n=2m+1$
$\Rightarrow 2m=n-1$
So, $S_{2m}=\frac{(n-1){(n-1+1)}^2}{2}$
$\Rightarrow S_{2m}=\frac{(n-1)n^2}{2}$ $...\left(2\right)$
Also, the odd terms of the given series are $3^2,5^2,...etc$.
$\Rightarrow {(2m+1)}^{th}term={(2m+1)}^2$
$\Rightarrow {(2m+1)}^{th}term=n^2$ $...\left(3\right)$
Now, substitute $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$; we get
$S_{2m+1}=\frac{(n-1)n^2}{2}+n^2$

$\Rightarrow S_{2m+1}=\frac{n^3-n^2+2n^2}{2}$

$\Rightarrow S_{2m+1}=\frac{n^3+n^2}{2}$

$\Rightarrow S_{2m+1}=\frac{n^2(n+1)}{2}$
If $n=4l+3$, then
$S_{2m+1}=\frac{(4l+3{)}^2(4l+4)}{2}$, which is even.
Hence, options 'A' and 'D' are correct.