Question

$\frac{1^3}{1}+\frac{(1^3+2^3)}{(1+2)}+\frac{(1^3+2^3+3^3)}{(1+2+3)}+....15terms=$

• A. $446$
• B. $680$
• C. $600$
• D. $540$

Answer 1

The correct option is B

$680$

Explanation of the correct option:

Step 1. Find the $n^{th}$ of given series:

$\Rightarrow \frac{1^3}{1}+\frac{(1^3+2^3)}{(1+2)}+\frac{(1^3+2^3+3^3)}{(1+2+3)}+....$

Here, $nth$ term of given series is

$a_n=\frac{1^3+2^3+3^3+....+n^3}{1+2+3+....+n}$

Step 2. Use sum of square and cubes of first $n$natural number formulas

Numerator of $a_n$:

$\Rightarrow 1^3+2^3+3^3+....+n^3$

$\Rightarrow \sum _{n=1}^n{n}^3={\left(\frac{n(n+1)}{2}\right)}^2$ ………(1)

Denominator of $a_n$:

$\Rightarrow 1+2+3+....+nterms$

$\Rightarrow \sum _{}n=\frac{n}{2}\left[n+1\right]=\frac{n(n+1)}{2}$ …….(2)

Substitute the values of equation (1) and (2) in $a_n$:

$\Rightarrow a_n=\frac{{\left({\displaystyle \frac{n(n+1)}{2}}\right)}^2}{\left({\displaystyle \frac{n(n+1)}{2}}\right)}=\frac{n(n+1)}{2}$

Step 3. Find the sum of $n$ terms of series:

$\Rightarrow S_n=\sum _{n=1}^n{a}_n=\sum _{n=1}^n\frac{n(n+1)}{2}$

$\Rightarrow \frac{1}{2}\left(\underset{n=1}{\overset{n}{\sum n^2}}+\underset{n=1}{\overset{n}{\sum n}}\right)$

$\Rightarrow \frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right)$

$\Rightarrow \frac{1}{2}n\left(\frac{(n+1)(2n+1)+3(n+1)}{6}\right)$

$\Rightarrow \frac{n}{12}\left(2n^2+n+2n+3n+1+3\right)$

$\Rightarrow \frac{n}{12}\left(2n^2+6n+4\right)$

Step 4. Substitute $n=15$ in above equation:

$S_{15}=\frac{15}{12}\left[2{\left(15\right)}^2+6\left(15\right)+4\right]$

$=\frac{544\times 15}{12}=\frac{8160}{12}=\mathbf{680}$

Hence, Option ‘B’ is Correct.