Question

The sum $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots$ to 16 terms is

• A. $246$
• B. $646$
• C. $446$
• D. $746$

Answer 1

Answer: (C)

$446$

$T_n=\frac{1^3+2^3+3^3+\cdots +n^3}{1+3+5+\dots nterms}=\frac{{\sum }^3}{\frac{n}{2}[2\times 1+(n-1\left)2\right]}$

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

Now,

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

$=\frac{1}{4}[\frac{n(n+1)(2n+1)}{6}+2\times \frac{n(n+1)}{2}+n]$

$=\frac{n}{24}[2n^2+3n+1+6n+6+6]$

$\frac{n}{24}[2n^2+9n+13]$

Putting, n = 16, we get

$S_{16}=\frac{16}{24}\left[2\right(256)+144+13]$

$=\frac{2}{3}\left(669\right)=446$