Question

Find the sum of \[\dfrac{{{1^3}}}{1} + \dfrac{{{1^3} + {2^3}}}{{1 + 2}} + \dfrac{{{1^3} + {2^3} + {3^3}}}{{1 + 2 + 3}} + ......................................16{\text{ terms}}\]
This question has multiple correct options
A. \[123\]
B. \[\dfrac{{1785}}{4}\]
C. \[\dfrac{{1875}}{4}\]
D. \[234\]

Answer 1

Hint: In this question, first of all find the \[{n^{th}}\] term of the given sequence. Then find the summations of the \[n\] terms of the sequence. Then put \[n = 16\] to get the sum of the first 16 terms which is our required answer.

Complete step by step solution:
Given sequence is \[\dfrac{{{1^3}}}{1} + \dfrac{{{1^3} + {2^3}}}{{1 + 2}} + \dfrac{{{1^3} + {2^3} + {3^3}}}{{1 + 2 + 3}} + ......................................16{\text{ terms}}\]
Consider the \[{n^{th}}\] term of the sequence.
\[{T_n} = \dfrac{{{1^3} + {2^3} + {3^3} + .................................. + {n^3}}}{{1 + 3 + 5 + ............................ + \left( {2n - 1} \right)}}\]
We know that, \[{1^3} + {2^3} + {3^3} + .................................. + {n^3} = {\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2}\] and \[1 + 3 + 5 + ............................ + \left( {2n - 1} \right) = {n^2}\]. By, using these formulae we get
\[
   \Rightarrow {T_n} = \dfrac{{{{\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]}^2}}}{{{n^2}}} \\
   \Rightarrow {T_n} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{{4{n^2}}} \\
  \therefore {T_n} = \dfrac{1}{4}{\left( {n + 1} \right)^2} \\
\]
Here, the sum of the sequence is equal to the summation of \[{T_n}\] (\[n\] terms) i.e., \[\dfrac{{{1^3}}}{1} + \dfrac{{{1^3} + {2^3}}}{{1 + 2}} + \dfrac{{{1^3} + {2^3} + {3^3}}}{{1 + 2 + 3}} + ......................................n{\text{ terms}} = \sum {{T_n}} \]
\[
   \Rightarrow \sum {{T_n}} = \sum {\dfrac{{{{\left( {n + 1} \right)}^2}}}{4}} \\
   \Rightarrow \sum {{T_n}} = \dfrac{1}{4}\sum {{{\left( {n + 1} \right)}^2}} \\
\]
We know that, \[\sum {{{\left( {n + 1} \right)}^2}} = \dfrac{{\left( {n + 1} \right)\left( {n + 2} \right)\left( {2n + 3} \right)}}{6}\]. By using this formula, we get
\[
   \Rightarrow \sum {{T_n}} = \dfrac{1}{4}\sum {{{\left( {n + 1} \right)}^2}} = \dfrac{1}{4} \times \dfrac{{\left( {n + 1} \right)\left( {n + 2} \right)\left( {2n + 3} \right)}}{6} \\
  \therefore \sum {{T_n}} = \dfrac{1}{4}\sum {{{\left( {n + 1} \right)}^2}} = \dfrac{{\left( {n + 1} \right)\left( {n + 2} \right)\left( {2n + 3} \right)}}{{24}} \\
\]
Since, there are 16 terms in the given sequence put \[n = 16\].
\[
   \Rightarrow \sum {{T_{16}} = \dfrac{{\left( {16 + 1} \right)\left( {16 + 2} \right)\left( {2 \times 16 + 3} \right)}}{{24}}} \\
   \Rightarrow \sum {{T_{16}} = \dfrac{{17 \times 18 \times 35}}{{24}}} \\
  \therefore \sum {{T_{16}} = \dfrac{{10710}}{{24}} = \dfrac{{1785}}{4}} \\
\]
Therefore, \[\dfrac{{{1^3}}}{1} + \dfrac{{{1^3} + {2^3}}}{{1 + 2}} + \dfrac{{{1^3} + {2^3} + {3^3}}}{{1 + 2 + 3}} + ......................................16{\text{ terms}} = \dfrac{{1785}}{4}\].
Thus, the correct option is B. \[\dfrac{{1785}}{4}\]

Note: Here, we have used the formula of sum of cubes of first \[n\] natural numbers i.e., \[{1^3} + {2^3} + {3^3} + .................................. + {n^3} = {\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2}\] and the sum of first \[n\] odd numbers i.e., \[1 + 3 + 5 + ............................ + \left( {2n - 1} \right) = {n^2}\].