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

Answer 1

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

The correct options are
A

$9^{th}$ term is 25

B

$7^{th}$ term is 16

C

The sum of first ten terms is $\frac{505}{4}$.

$\therefore T_r=\frac{1}{r^2}(1+\mathrm{2...}+r{)}^2$

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

$\therefore T_7=16$

$T_9=25$

$S_{10}=\frac{{\sum }_{r=1}^{10}{r}^2+2{\sum }_{r=1}^{10}r+{\sum }_{r=1}^{10}1}{4}$

$=\frac{1}{4}[\frac{10\times 11\times 21}{6}+2.\frac{10\times 11}{2}+10]$

$=\frac{505}{4}$