Question

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

• A. $7$th term is $16$
• B. $7$th term is $18$
• C. Sum of first $10$ terms is $\frac{505}{4}$
• D. Sum of first $10$ terms is $\frac{405}{4}$

Answer 1

Answer: (A), (C)

Sum of first $10$ terms is $\frac{505}{4}$

$S=1+\frac{1}{(1+3)}(1+2{)}^2+\frac{1}{(1+3+5)}(1+2+3{)}^2+\frac{1}{(1+3+5+7)}(1+2+3+4{)}^2+\dots \dots$
The $r^{th}$ term is given by
$T_r=\frac{1}{r^2}(1+2+\dots +r{)}^2$
$=\frac{1}{r^2}{\left\{\frac{r(r+1)}{2}\right\}}^2$
$=\frac{r^2+2r+1}{4}$
$\therefore T_7=16$

$S_{10}=\sum _{r=1}^{10}{T}_r$
$=\frac{1}{4}\left(\sum _{r=1}^{10}{r}^2+2\sum _{r=1}^{10}r+\sum _{r=1}^{10}1\right)$
$=\frac{1}{4}(\frac{\left(10\right)(10+1)(20+1)}{6}+(10\left)\right(10+1)+10)=\frac{505}{4}$