Question

Find the sum of the series
$\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+.........+n$

• A. Sum of $n$ terms $S_n=\frac{n}{12}(2n^2+9n+13)$
• B. Sum of $n$ terms $S_n=\frac{n}{9}(2n^3+9n+13)$
• C. Sum of $n$ terms $S_n=\frac{n}{24}(2n^3+9n+13)$
• D. Sum of $n$ terms $S_n=\frac{n}{24}(2n^2+9n+13)$

Answer 1

The correct option is D Sum of $n$ terms $S_n=\frac{n}{24}(2n^2+9n+13)$

Here $n$th term $T_n=\frac{1^3+2^3+3^3+.....+n^3}{1+3+5+.....+(2n-1)}$
$=\frac{\sum n^3}{\frac{n}{2}(1+2n-1)}$
$=\frac{{\left[n\left(\frac{n+1}{2}\right)\right]}^2}{n^2}$
$=\frac{1}{4}(n+1{)}^2$
$=\frac{1}{4}(n^2+2n+1)$
$=\frac{1}{4}{n}^2+\frac{1}{2}n+\frac{1}{4}$
$\therefore$ Sum of $n$ terms,
$S_n=\sum T_n$
$=\frac{1}{4}\sum n^2+\frac{1}{2}\sum n+\frac{1}{4}\sum n$
$=\frac{1}{4}.\frac{n(n+)(2n+1)}{6}+\frac{1}{2}.\frac{n(n+1)}{2}+\frac{1}{4}.n$
$=\frac{1}{24}n(n+1)(2n+1)+\frac{1}{4}n(n+1)+\frac{1}{4}n$
$=\frac{n}{24}\{2n^2+3n+1+6n+6+6\}$
$S_n=\frac{n}{24}(2n^2+9n+13)$