Question

Find the sum of the series \[\dfrac{{{1^3}}}{1}{\text{ }} + {\text{ }}\dfrac{{\left( {{1^3} + {2^3}} \right)}}{2}{\text{ }} + {\text{ }}\dfrac{{\left( {{1^3} + {2^3} + {3^3}} \right)}}{3}{\text{ }} + ...\] up to n terms is equal to
A. \[\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {5n + 3} \right)}}{{48}}\]
B. \[\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{{24}}\]
C. \[\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {7n + 1} \right)}}{{48}}\]
D. \[\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 5} \right)}}{{48}}\]

Answer 1

Hint: To solve this question, we need to find the \[{n_{th}}\] term of the numerator and the denominator separately then find the sum of the series by using the sum of squares on \[n\] natural numbers, the sum of cubes of \[n\] natural numbers and the sum of \[n\] numbers.

Formula Used:
1. \[{T_n} = a + \left( {n - 1} \right)d\]
where \[a\] is the first term, \[d\] is a common difference and \[{T_n}\] is \[{n_{th}}\] term of series
2. \[{S_n} = \sum {{T_n}} \]
where \[{S_n}\] is the sum of the series and \[{T_n}\] is the \[{n_{th}}\] of the series
3. \[\sum {{n^3} = } \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]
4. \[\sum {{n^2} = } {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}\]
5. \[\sum {n = } \dfrac{{n\left( {n + 1} \right)}}{2}\]

Complete step-by-step solution:
Given that \[\dfrac{{{1^3}}}{1}{\text{ }} + {\text{ }}\dfrac{{\left( {{1^3} + {2^3}} \right)}}{2}{\text{ }} + {\text{ }}\dfrac{{\left( {{1^3} + {2^3} + {3^3}} \right)}}{3}{\text{ }} + ...upto{\text{ }}n{\text{ }}terms{\text{ }}\]
Now solve the \[{n_{th}}\] term of the numerator and the denominator separately:
Now the numerator series is \[{1^3} + {2^3} + {3^3}... + {n^3}\]
Now, the \[{n_{th}}\] term of the series by formula \[{T_n} = a + \left( {n - 1} \right)d\] is
\[
  {T_n} = {1^3} + {2^3} + ..... + { n^3} \\
   = \sum {{n^3}}
 \]
We know that sum of cubes of n natural number is \[{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}\]
Therefore,
\[
  {T_n} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} \\
   = \dfrac{{{{\left[ {n\left( {n + 1} \right)} \right]}^2}}}{{{2^2}}} \\
   = \dfrac{{{{\left[ {n\left( {n + 1} \right)} \right]}^2}}}{4}...\left( 1 \right)
 \]
Now the denominator series is \[1,2,3,...n\]
Now, the \[{n_{th}}\] term of the series by formula \[{T_n} = a + \left( {n - 1} \right)d\] is
\[
  {T_n} = a + \left( {n - 1} \right)d \\
   = 1 + \left( {n - 1} \right)1 \\
   = 1 + n - 1 \\
   = n...\left( 2 \right)
 \]
Therefore, the \[{n_{th}}\] of the series \[\dfrac{{{1^3}}}{1}{\text{ }} + {\text{ }}\dfrac{{\left( {{1^3} + {2^3}} \right)}}{2}{\text{ }} + {\text{ }}\dfrac{{\left( {{1^3} + {2^3} + {3^3}} \right)}}{3}{\text{ }} + ...upto{\text{ }}n{\text{ }}terms{\text{ }}\] from equation (1) and (2) is
\[
  {T_n} = \dfrac{{\dfrac{{{{\left[ {n\left( {n + 1} \right)} \right]}^2}}}{4}}}{n} \\
   = \dfrac{{{{\left[ {n\left( {n + 1} \right)} \right]}^2}}}{{4n}} \\
   = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{{4n}} \\
   = \dfrac{{n{{\left( {n + 1} \right)}^2}}}{4}
 \]
Now, the sum of series is given by \[{S_n} = \sum {{T_n}} \]
\[
  {S_n} = \sum {\dfrac{{n{{\left( {n + 1} \right)}^2}}}{4}} \\
   = \sum {\dfrac{{n\left( {{n^2} + 1 + 2n} \right)}}{4}} \\
   = \sum {\dfrac{{{n^3} + n + 2{n^2}}}{4}} \\
   = \dfrac{1}{4}\sum {{n^3} + n + 2{n^2}}
 \]
Further solving,
\[{S_n} = \dfrac{1}{4}\left[ {\sum {{n^3} + \sum {2{n^2}} } + \sum n } \right]\]
We know that \[\sum {{n^3}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\], \[\sum {{n^2}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}\]and \[\sum n = \dfrac{{n\left( {n + 1} \right)}}{2}\]
Now by applying these formulas, we get
\[
  {S_n} = \dfrac{1}{4}\left[ {{{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)}^2} + 2\left( {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) + \dfrac{{n\left( {n + 1} \right)}}{2}} \right] \\
   = \dfrac{1}{4}\left[ {\dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{{{2^2}}} + \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{3} + \dfrac{{n\left( {n + 1} \right)}}{2}} \right] \\
   = \dfrac{1}{4}\left[ {n\left( {n + 1} \right)\left[ {\dfrac{{n\left( {n + 1} \right)}}{4} + \dfrac{{2n + 1}}{3} + \dfrac{1}{2}} \right]} \right] \\
   = \dfrac{{n\left( {n + 1} \right)}}{4}\left[ {\dfrac{{3\left( {n\left( {n + 1} \right)} \right) + 4\left( {2n + 1} \right) + 6}}{{12}}} \right]
 \]
Further solving,
\[
  {S_n} = \dfrac{{n\left( {n + 1} \right)}}{4}\left[ {\dfrac{{3\left( {{n^2} + n} \right) + 8n + 4 + 6}}{{12}}} \right] \\
   = \dfrac{{n\left( {n + 1} \right)}}{4}\left[ {\dfrac{{3{n^2} + 3n + 8n + 10}}{{12}}} \right] \\
   = \dfrac{{n\left( {n + 1} \right)}}{4}\left[ {\dfrac{{3{n^2} + 11n + 10}}{{12}}} \right] \\
   = \dfrac{{n\left( {n + 1} \right)}}{4}\left[ {\dfrac{{3{n^2} + 6n + 5n + 10}}{{12}}} \right]
 \]
Furthermore solving,
\[
  {S_n} = \dfrac{{n\left( {n + 1} \right)}}{4}\left[ {\dfrac{{3n\left( {n + 2} \right) + 5\left( {n + 2} \right)}}{{12}}} \right] \\
   = \dfrac{{n\left( {n + 1} \right)}}{4}\left[ {\dfrac{{\left( {3n + 5} \right)\left( {n + 2} \right)}}{{12}}} \right] \\
   = \dfrac{{n\left( {n + 1} \right)\left( {3n + 5} \right)\left( {n + 2} \right)}}{{4 \times 12}} \\
   = \dfrac{{n\left( {n + 1} \right)\left( {3n + 5} \right)\left( {n + 2} \right)}}{{48}}
 \]

Hence, option (D) is correct

Note: Using the series formula, we calculated the sum of the given series. Students must be careful while calculating the sum of series as he/she could be mistakes while calculating the series and apply the formulas used in the above solution correctly to get the required result.