Question

Sum of the following series up to n terms is
1.2.3.4 + 2.3.4.5+ 3.4.5.6 +.................

Answer 1

Hint: In this particular question first find out the $ {k^{th}} $ term of the given series, then apply summation in the $ {k^{th}} $ term from 1 to n, then apply direct general formula of the summation such as, $ \sum\limits_{k = 1}^n k = \dfrac{{n\left( {n + 1} \right)}}{2},\sum\limits_{k = 1}^n {{k^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} $ and so on, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given series is,
1.2.3.4 + 2.3.4.5+ 3.4.5.6 +.................
We have to find out the sum of this series.
So first find out the general term of this series, i.e. $ {k^{th}} $ term.
So when we observe carefully the $ {k^{th}} $ term is given as, k (K + 1) (k + 2) (k + 3)
Now apply summation so the sum of the given series is written as,
 $ \Rightarrow {S_n} = \sum\limits_{k = 1}^n {k\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)} $
Now simplify this we have,
 $ \Rightarrow {S_n} = \sum\limits_{k = 1}^n {\left( {{k^2} + k} \right)\left( {{k^2} + 5k + 6} \right)} $
 $ \Rightarrow {S_n} = \sum\limits_{k = 1}^n {\left( {{k^4} + 5{k^3} + 6{k^2} + {k^3} + 5{k^2} + 6k} \right)} $
 $ \Rightarrow {S_n} = \sum\limits_{k = 1}^n {\left( {{k^4} + 6{k^3} + 11{k^2} + 6k} \right)} $
Now separate its terms we have,
 $ \Rightarrow {S_n} = \sum\limits_{k = 1}^n {{k^4}} + 6\sum\limits_{k = 1}^n {{k^3}} + 11\sum\limits_{k = 1}^n {{k^2}} + 6\sum\limits_{k = 1}^n k $
Now as we know that, \[\sum\limits_{k = 1}^n k = \dfrac{{n\left( {n + 1} \right)}}{2},\sum\limits_{k = 1}^n {{k^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6},\sum\limits_{k = 1}^n {{k^3}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}\]
And, $ \sum\limits_{k = 1}^n {{k^4}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)\left( {3{n^2} + 3n - 1} \right)}}{{30}} $ now substitute these values in the above equation we have,
 $ \Rightarrow {S_n} = 6\dfrac{{n\left( {n + 1} \right)}}{2} + 11\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + 6{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} + \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)\left( {3{n^2} + 3n - 1} \right)}}{{30}} $
 $ \Rightarrow {S_n} = 3{n^2} + 3 + 11\dfrac{{\left( {{n^2} + n} \right)\left( {2n + 1} \right)}}{6} + 6\left( {\dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}} \right) + \dfrac{{\left( {{n^2} + n} \right)\left( {2n + 1} \right)\left( {3{n^2} + 3n - 1} \right)}}{{30}} $
 $ \Rightarrow {S_n} = 3{n^2} + 3 + 11\dfrac{{\left( {2{n^3} + {n^2} + 2{n^2} + n} \right)}}{6} + 6\left( {\dfrac{{{n^2}\left( {{n^2} + 1 + 2n} \right)}}{4}} \right) + \dfrac{{\left( {2{n^3} + {n^2} + 2{n^2} + n} \right)\left( {3{n^2} + 3n - 1} \right)}}{{30}} $
 $ \Rightarrow {S_n} = 3{n^2} + 3 + 11\dfrac{{\left( {2{n^3} + 3{n^2} + n} \right)}}{6} + 6\dfrac{{\left( {{n^4} + {n^2} + 2{n^3}} \right)}}{4} + \dfrac{{\left( {2{n^3} + 3{n^2} + n} \right)\left( {3{n^2} + 3n - 1} \right)}}{{30}} $
 $ \Rightarrow {S_n} = 3{n^2} + 3 + \dfrac{{11}}{3}{n^3} + \dfrac{{11}}{2}{n^2} + 11\dfrac{n}{6} + \dfrac{3}{2}{n^4} + \dfrac{3}{2}{n^2} + 3{n^3} + \dfrac{{\left( {6{n^5} + 15{n^4} + 10{n^3} - n} \right)}}{{30}} $
 $ \Rightarrow {S_n} = 3{n^2} + 3 + \dfrac{{11}}{3}{n^3} + \dfrac{{11}}{2}{n^2} + 11\dfrac{n}{6} + \dfrac{3}{2}{n^4} + \dfrac{3}{2}{n^2} + 3{n^3} + \dfrac{1}{5}{n^5} + \dfrac{1}{2}{n^4} + \dfrac{1}{3}{n^3} - \dfrac{1}{{30}}n $
 $ \Rightarrow {S_n} = \dfrac{1}{5}{n^5} + \dfrac{1}{2}{n^4} + \dfrac{3}{2}{n^4} + \dfrac{{11}}{3}{n^3} + 3{n^3} + \dfrac{1}{3}{n^3} + 3{n^2} + \dfrac{{11}}{2}{n^2} + \dfrac{3}{2}{n^2} + \dfrac{{11}}{6}n - \dfrac{1}{{30}}n + 3 $
 $ \Rightarrow {S_n} = \dfrac{1}{5}{n^5} + {n^4}\left( {\dfrac{1}{2} + \dfrac{3}{2}} \right) + {n^3}\left( {\dfrac{{11}}{3} + 3 + \dfrac{1}{3}} \right) + {n^2}\left( {3 + \dfrac{{11}}{2} + \dfrac{3}{2}} \right) + n\left( {\dfrac{{11}}{6} - \dfrac{1}{{30}}} \right) + 3 $
 $ \Rightarrow {S_n} = \dfrac{1}{5}{n^5} + 2{n^4} + 7{n^3} + 10{n^2} + \dfrac{9}{5}n + 3 $
So this is the required sum of the given series.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formulas of the sum of first natural numbers, square of first natural numbers, cube of first natural numbers, fourth power of first natural numbers which is all stated above so simply substitute then and simplify we will get the required sum of the given series.