Question

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

• A. $9^{\text{th}}$ term is $25$
• B. $7^{\text{th}}$ term is $16$
• C. Sum of first ten terms is $\frac{505}{4}$
• D. Sum of first ten terms is $505$

Answer 1

Answer: (A), (B), (C)

Sum of first ten terms is $\frac{505}{4}$

We have, $S=1+\frac{1}{(1+3)}(1+2{)}^2+\frac{1}{(1+3+5)}(1+2+3{)}^2+\frac{1}{(1+3+5+7)}(1+2+3+4{)}^2+\cdots$
So, the general term can be written as
$T_r=\frac{1}{r^2}(1+2+\cdots +r{)}^2=\frac{1}{r^2}{\left(\frac{r(r+1)}{2}\right)}^2$
$=\frac{(r+1{)}^2}{4}=\frac{r^2+2r+1}{4}$
$\therefore T_7=16$
And, $T_9=25$

$S_{10}=\frac{\sum _{r=1}^{10}{r}^2+2\sum _{r=1}^{10}r+\sum _{r=1}^{10}1}{4}$
$=\frac{1}{4}[\frac{10\times 11\times 21}{6}+2\left(\frac{10\times 11}{2}\right)+10]$
$=\frac{505}{4}$