Question

Find the sum of the given expression:
$3 \times {1^2} + 5 \times {2^2} + 7 \times {3^2} + .....$

Answer 1

Hint: In this question, first we find the nth term of the given expression. Now to find the sum we will use the sigma notation. We will apply the sigma operator on each term and then finally get the sum. We will use the summation results: $\sum\limits_{k = 1}^n {{k^3}} = {\left( {\dfrac{{n(n + 1)}}{2}} \right)^2}{\text{ and }}\sum\limits_{k = 1}^n {{k^2}} = \left( {\dfrac{{n(n + 1)(2n + 1)}}{6}} \right){\text{ }}$

Complete step-by-step answer:
The given expression is:
$3 \times {1^2} + 5 \times {2^2} + 7 \times {3^2} + .....$ and we have to find the summation result of this expression.
Here the terms are neither in GP nor in AP. Also the terms are not in AGP. So, the method that is left to find the summation is using the sigma operator.
Here we will first write the nth term of the given expression and then apply the sigma operator.
In each of the terms, we have an odd number and another is a square of natural numbers starting from 1.
The general term for odd numbers is (2n+1).
Therefore, the nth term of the given expression is written as:
${T_n} = (2n + 1){n^2}$.
To find the sum of all the terms, we use the sigma operator.
$\therefore $ ${S_{_n}} = \sum\limits_{n = 1}^n {{T_n}} $
Putting the value of ${T_n}$, we have:
${S_{_n}} = \sum\limits_{n = 1}^n {(2n + 1){n^2}} = \sum\limits_{n = 1}^n {2{n^3}} + \sum\limits_{n = 1}^n {{n^2}} $
Now, we know that:
$\sum\limits_{k = 1}^n {{k^3}} = {\left( {\dfrac{{n(n + 1)}}{2}} \right)^2}{\text{ and }}\sum\limits_{k = 1}^n {{k^2}} = \left( {\dfrac{{n(n + 1)(2n + 1)}}{6}} \right){\text{ }}$.
Using these results, we can write our original summation as:
${S_{_n}} = 2{\left( {\dfrac{{n(n + 1)}}{2}} \right)^2}{\text{ + }}\left( {\dfrac{{n(n + 1)(2n + 1)}}{6}} \right)$
\[ \Rightarrow {S_n} = \dfrac{{{n^2}{{(n + 1)}^2}}}{2}{\text{ + }}\left( {\dfrac{{n(n + 1)(2n + 1)}}{6}} \right)\]
On taking the $\dfrac{{n(n + 1)}}{2}$ term common, we get:
${S_{_n}} = \dfrac{{n(n + 1)}}{2}\left[ {{\text{n(n + 1) + }}\dfrac{{2n + 1}}{3}} \right]$
On further solving, we get:
${S_{_n}} = \dfrac{{n(n + 1)}}{2}\left[ {\dfrac{{3{n^2} + 5n + 1}}{3}} \right]$
$ \Rightarrow {S_{_n}} = \dfrac{{n(n + 1)(3{n^2} + 5n + 1)}}{6}$ .
Therefore, the sum of the given expression is ${S_{_n}} = \dfrac{{n(n + 1)(3{n^2} + 5n + 1)}}{6}$.

Note: In this type of question, where the terms are not seen to be in AP or GP or any other progression, then in such case it is better to write the nth term and then use the sigma operator to find the sum of series. You should know the following theorems related to sigma notation:
1.$\sum\limits_{r = 1}^n {\left( {{a_r} \pm {b_r}} \right)} = \sum\limits_{r = 1}^n {{a_r}} \pm \sum\limits_{r = 1}^n {{b_r}} $
2.$\sum\limits_{r = 1}^n {k{a_r}} = k\sum\limits_{r = 1}^n {{a_r}} {\text{ and }}\sum\limits_{r = 1}^n k = nk$; where k is a constant.