Question

For the series, $S=1+\frac{1}{(1+3)}(1+2{)}^2+\frac{1}{(1+3+5)}(1+2+3{)}^2+\frac{1}{(1+3+5+7)}(1+2+3+4{)}^2+....$

• A. $7\text{th}$ term is $16$
• B. $11\text{th}$ term is $32$
• C. sum of first $7$ terms is $\frac{203}{4}$
• D. sum of first $10$ terms is $\frac{505}{4}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $7\text{th}$ term is $16$
C sum of first $7$ terms is $\frac{203}{4}$
D sum of first $10$ terms is $\frac{505}{4}$

$S=1+\frac{1}{(1+3)}(1+2{)}^2+\frac{1}{(1+3+5)}(1+2+3{)}^2+\frac{1}{(1+3+5+7)}(1+2+3+4{)}^2+....$

$T_n=\frac{1}{n^2}(1+2+...+n{)}^2$
$=\frac{1}{n^2}{\left(\frac{n(n+1)}{2}\right)}^2$
$=\frac{n^2+2n+1}{4}$

$T_7=16,T_{11}=36$

$S_n=\sum T_n$
$=\frac{1}{4}(\sum n^2+2\sum n+\sum 1)$
$=\frac{1}{4}[\frac{n(n+1)(2n+1)}{6}+n(n+1)+n]$
$=\frac{n}{24}\left[\right(n+1\left)\right(2n+7)+6]$

$S_7=\frac{203}{4},S_{10}=\frac{505}{4}$