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., $1 + 3 + 5 + ............................ + \left( {2n - 1} \right) = {n^2}$.