Question

The sum of n terms of the series $1 \cdot 2 \cdot 5 + 2 \cdot 3 \cdot 6 + 3 \cdot 4 \cdot 7 + ..........n{\text{ terms}}$ is

Answer 1

Hint- To evaluate the sum of n terms first determine the general terms of the given series. Then proceed further by putting the value of general term in the formula of sum of n terms which is given by:
${S_n} = \sum\limits_{n = 1}^n {{T_n}} $
${S_n} = $ sum of n terms of the series
${T_n} = $ nth term of the series

Complete step-by-step solution -
Given series is $1 \cdot 2 \cdot 5 + 2 \cdot 3 \cdot 6 + 3 \cdot 4 \cdot 7 + ..........n{\text{ terms}}$
We will try to find the general term with the help of given terms
As the first number of each term is the same as the term number.
So let us denote the first number of nth term as n
Second number of each term is 1 more than the first number of each term, so the second number can be denoted by (n+1)
Third number of each term is 4 more than the first number of each term, so the third number can be denoted by (n+4)
So, the general term is:
\[
  {T_n} = n\left( {n + 1} \right)\left( {n + 4} \right) \\
   \Rightarrow {T_n} = n\left( {{n^2} + 4n + n + 4} \right) \\
   \Rightarrow {T_n} = n\left( {{n^2} + 5n + 4} \right) \\
   \Rightarrow {T_n} = {n^3} + 5{n^2} + 4n \\
 \]
As now we have found out the general term, let us move on to finding the sum of n terms.
We know that the sum of n terms of the series is given by ${S_n} = \sum\limits_{n = 1}^n {{T_n}} $
So by putting the value ${T_n}$ in the above formula, we obtain
\[
  {S_n} = \sum\limits_{n = 1}^n {{T_n}} \\
   \Rightarrow {S_n} = \sum\limits_{n = 1}^n {\left( {{n^3} + 5{n^2} + 4n} \right)} \\
   \Rightarrow {S_n} = \sum\limits_{n = 1}^n {\left( {{n^3}} \right)} + \sum\limits_{n = 1}^n {\left( {5{n^2}} \right)} + \sum\limits_{n = 1}^n {\left( {4n} \right)} \\
   \Rightarrow {S_n} = \sum\limits_{n = 1}^n {\left( {{n^3}} \right)} + 5\sum\limits_{n = 1}^n {\left( {{n^2}} \right)} + 4\sum\limits_{n = 1}^n {\left( n \right)} \\
 \]
As we know the general for sum of n terms belonging to series $n,{n^2}\& {n^3}$ which is given by:
\[
  \sum\limits_{n = 1}^n {\left( {{n^3}} \right)} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} \\
  \sum\limits_{n = 1}^n {\left( {{n^2}} \right)} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \\
  \sum\limits_{n = 1}^n {\left( n \right)} = \dfrac{{n\left( {n + 1} \right)}}{2} \\
 \]
Using these general formulas to find the sum of the n terms of the series
\[
   \Rightarrow {S_n} = \sum\limits_{n = 1}^n {\left( {{n^3}} \right)} + 5\sum\limits_{n = 1}^n {\left( {{n^2}} \right)} + 4\sum\limits_{n = 1}^n {\left( n \right)} \\
   \Rightarrow {S_n} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + 5 \times \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + 4 \times \dfrac{{n\left( {n + 1} \right)}}{2} \\
 \]
Simplifying the terms by taking some terms common and taking LCM we get
\[
   \Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{n\left( {n + 1} \right)}}{2} + \dfrac{{5\left( {2n + 1} \right)}}{3} + 4} \right] \\
   \Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{3n\left( {n + 1} \right) + 10\left( {2n + 1} \right) + 24}}{6}} \right] \\
   \Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{3{n^2} + 3n + 20n + 10 + 24}}{6}} \right] \\
   \Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{3{n^2} + 23n + 34}}{6}} \right] \\
   \Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)\left( {3{n^2} + 23n + 34} \right)}}{{12}} \\
 \]
Hence, the sum of n terms of the given series is \[\dfrac{{n\left( {n + 1} \right)\left( {3{n^2} + 23n + 34} \right)}}{{12}}\] .

Note- In order to find the sum of such series which does not belong to any specified series and contains some random number, always try to establish some relationship between the terms to find the general equation of nth term and finally use the formula of sum of n terms for general series. Always remember the sum of n terms of some common series such as $n,{n^2},{n^3}...{\text{etc}}$ . They are very useful in solving such questions.