Question

The sum of the given series is
$1 + \dfrac{{{1^3} + {2^3}}}{{1 + 2}} + \dfrac{{{1^3} + {2^3} + {3^3}}}{{1 + 2 + 3}} + .......... + \dfrac{{{1^3} + {2^3} + {3^3} + ...... + {{15}^3}}}{{1 + 2 + 3 + ...... + 15}} - \dfrac{1}{2}\left( {1 + 2 + 3 + ...... + 15} \right)$
A) 1240
B) 1860
C) 660
D) 620

Answer 1

Hint: Here we bring the given series in summation form and solve them.

Complete step-by-step answer:
The given series is written as
$1 + \dfrac{{{1^3} + {2^3}}}{{1 + 2}} + \dfrac{{{1^3} + {2^3} + {3^3}}}{{1 + 2 + 3}} + .......... + \dfrac{{{1^3} + {2^3} + {3^3} + ...... + {{15}^3}}}{{1 + 2 + 3 + ...... + 15}} - \dfrac{1}{2}\left( {1 + 2 + 3 + ...... + 15} \right)$
Now it is written as
$\sum\limits_{r = 1}^{15} {\dfrac{{{1^3} + {2^3} + {3^3} + ...... + {{15}^3}}}{{1 + 2 + 3 + ...... + 15}}} - \dfrac{1}{2}\sum\limits_{r = 1}^{15} r $
Now you know
${\sum\limits_{r = 1}^n {{r^2} = \left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)} ^2}$ And $\sum\limits_{r = 1}^n {r = \left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)} $
In this question n=15, so apply this
\[\left[ {{{\sum\limits_{r = 1}^{15} {\dfrac{{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)}}{{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)}}} }^2}} \right] - \dfrac{1}{2}\left[ {\dfrac{{15(16)}}{2}} \right]\]
\[\left[ {\sum\limits_{r = 1}^{15} {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)} } \right] - 60\]
Now break the summation
\[\left[ {\dfrac{1}{2}\sum\limits_{r = 1}^{15} {{n^2} + \dfrac{1}{2}\sum\limits_{r = 1}^{15} n } } \right] - 60\]
Now you know that $\sum\limits_{r = 1}^n {{r^3}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
$ \Rightarrow \dfrac{1}{2}\left[ {\dfrac{{15(16)(31)}}{6}} \right] + \dfrac{1}{2}\left[ {\dfrac{{15(16)}}{2}} \right] - 60$
$ \Rightarrow $ 620 + 60 – 60
$ \Rightarrow $620
So option D is correct.

Note: In this type of problem remember the summation formulas it will help you a lot in solving these types of series.