Question

If the nth term of a series be $ 3 + n\left( {n - 1} \right) $ , then the sum of n terms of the series is:
(A) $ \dfrac{{{n^2} + n}}{3} $
(B) $ \dfrac{{{n^3} + 8n}}{3} $
(C) $ \dfrac{{{n^2} + 8n}}{5} $
(D) $ \dfrac{{{n^2} - 8n}}{3} $

Answer 1

Hint: In the given question, we are provided with the nth term of a series and are required to find the sum of n terms. So, we should know the formula for sum of first n natural numbers and sum of squares of first n natural numbers to get to the correct answer of the given problem. Algebraic and simplification rules can be used to simplify and ease the calculations.

Complete step-by-step answer:
So, we have, $ {a_n} = 3 + n\left( {n - 1} \right) $
 $ \Rightarrow {a_n} = {n^2} - n + 3 $
Finding the summation of n terms of this series. So, we get,
 $ {S_n} = \sum {{n^2} - n + 3} $
Distributing the summation on all the terms separately, we get,
 $ \Rightarrow {S_n} = \sum {{n^2}} - \sum n + \sum 3 $
Now, we know that multiplication is repeated addition. So, $ 3 $ added n times is equal to $ 3n $ . So, we get,
 $ \Rightarrow {S_n} = \sum {{n^2}} - \sum n + 3n $
Now, we know the formula for the sum of the first n natural numbers is \[\dfrac{{n\left( {n + 1} \right)}}{2}\]. So, substituting in the expression, we get,
 $ \Rightarrow {S_n} = \sum {{n^2}} - \dfrac{{n\left( {n + 1} \right)}}{2} + 3n $
Now, we also know that sum of squares of n natural numbers is \[\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]. Substituting this into the expression, we get,
 $ \Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \dfrac{{n\left( {n + 1} \right)}}{2} + 3n $
Now, we have to simplify the expression using simplification and algebraic rules.
So, taking out n common from all the terms, we get,
 $ \Rightarrow {S_n} = n\left[ {\dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \dfrac{{\left( {n + 1} \right)}}{2} + 3} \right] $
 $ \Rightarrow {S_n} = n\left[ {\dfrac{{2{n^2} + 3n + 1}}{6} - \dfrac{{\left( {n + 1} \right)}}{2} + 3} \right] $
Now, we take LCM of denominators.
\[ \Rightarrow {S_n} = n\left[ {\dfrac{{2{n^2} + 3n + 1 - 3n - 3 + 18}}{6}} \right]\]
Cancelling the like terms with same magnitude and opposite signs, we get,
\[ \Rightarrow {S_n} = n\left[ {\dfrac{{2{n^2} + 1 - 3 + 18}}{6}} \right]\]
\[ \Rightarrow {S_n} = n\left[ {\dfrac{{2{n^2} + 16}}{6}} \right]\]
Cancelling common factor in numerator and denominator, we get,
\[ \Rightarrow {S_n} = \left[ {\dfrac{{{n^3} + 8n}}{3}} \right]\]
So, the correct answer is option (B).
So, the correct answer is “Option B”.

Note: We must remember the formulae for summation of first n natural numbers and sum of squares of first n natural numbers in order to solve the problem. We must take the calculations and verify them so as to be sure of the final answer.