Question

If ${n^{th}}$ term of a series be $3 + n\left( {n - 1} \right)$. Then find the sum of $n$ terms of the series.
A. $\dfrac{{\left( {{n^2} + n} \right)}}{3}$
B. $\dfrac{{\left( {{n^3} + 8n} \right)}}{3}$
C. $\dfrac{{\left( {{n^2} + 8n} \right)}}{5}$
D. $\dfrac{{\left( {{n^2} - 8n} \right)}}{3}$

Answer 1

Hint: Here use the formula of sum of $n$ terms of the series. Substitute the given ${n^{th}}$ term of a series in the formula. Then divide the sum into sub parts. After that, apply the formulas of the sum of natural numbers and sum of squares of natural numbers to find the sum of $n$ terms of the series.

Formula Used:
The sum of natural numbers: $\sum\limits_{i = 1}^n i = \dfrac{{n\left( {n + 1} \right)}}{2}$
The sum of squares of natural numbers: $\sum\limits_{i = 1}^n {{i^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
The sum of $n$ terms of the series $ = \sum\limits_{n = 1}^n {{a_n}} $
$\sum\limits_{n = 1}^n 1 = n$

Complete step by step solution:
Given: The ${n^{th}}$ term of a series is $3 + n\left( {n - 1} \right)$.
Let consider ${a_n}$ be the ${n^{th}}$ term of a series.
Then,
${a_n} = 3 + n\left( {n - 1} \right)$
$ \Rightarrow {a_n} = 3 + {n^2} - n$
Apply the formula of the sum of $n$ terms of a series.
So, the sum of the $n$ terms in the given series is,
$Sum = \sum\limits_{n = 1}^n {{a_n}} $
Substitute the value of ${a_n}$ in the above formula.
$Sum = \sum\limits_{n = 1}^n {\left( {3 + {n^2} - n} \right)} $
Apply the summation on every term on right-hand side.
$Sum = \sum\limits_{n = 1}^n 3 + \sum\limits_{n = 1}^n {{n^2}} - \sum\limits_{n = 1}^n n $
Now use the formulas of the sum of natural numbers and sum of squares of natural numbers.
$Sum = 3\sum\limits_{n = 1}^n 1 + \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \dfrac{{n\left( {n + 1} \right)}}{2}$
$ \Rightarrow Sum = 3n + \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \dfrac{{n\left( {n + 1} \right)}}{2}$ $\because \sum\limits_{n = 1}^n 1 = n$
Simplify the above equation.
$Sum = n\left[ {3 + \dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \dfrac{{\left( {n + 1} \right)}}{2}} \right]$
Equalize the denominator of the terms present in the brackets.
$Sum = n\left[ {\dfrac{{18}}{6} + \dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \dfrac{{3\left( {n + 1} \right)}}{6}} \right]$
$ \Rightarrow Sum = \dfrac{n}{6}\left[ {18 + \left( {n + 1} \right)\left( {2n + 1} \right) - 3\left( {n + 1} \right)} \right]$
Simplify the above equation.
$Sum = \dfrac{n}{6}\left[ {18 + 2{n^2} + 2n + n + 1 - 3n - 3} \right]$
$ \Rightarrow Sum = \dfrac{n}{6}\left[ {2{n^2} + 16} \right]$
$ \Rightarrow Sum = \dfrac{{2n\left( {{n^2} + 8} \right)}}{6}$
$ \Rightarrow Sum = \dfrac{{n\left( {{n^2} + 8} \right)}}{3}$
$ \Rightarrow Sum = \dfrac{{\left( {{n^3} + 8n} \right)}}{3}$

Option ‘B’ is correct

Note: Use the summation method as we did above whenever you are given a series and asked to find the sum of the n terms. It would be simpler to use the formula for the sum of n terms of AP and GP instead of the summation if the given series is an AP or GP.