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Question: If the nth term of a series be \( 3 + n\left( {n - 1} \right) \) , then the sum of n terms of the se...

If the nth term of a series be 3+n(n−1)3 + n\left( {n - 1} \right) , then the sum of n terms of the series is:
(A) n2+n3\dfrac{{{n^2} + n}}{3}
(B) n3+8n3\dfrac{{{n^3} + 8n}}{3}
(C) n2+8n5\dfrac{{{n^2} + 8n}}{5}
(D) n2−8n3\dfrac{{{n^2} - 8n}}{3}

Explanation

Solution

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, an=3+n(n−1){a_n} = 3 + n\left( {n - 1} \right)
⇒an=n2−n+3\Rightarrow {a_n} = {n^2} - n + 3
Finding the summation of n terms of this series. So, we get,
Sn=∑n2−n+3{S_n} = \sum {{n^2} - n + 3}
Distributing the summation on all the terms separately, we get,
⇒Sn=∑n2−∑n+∑3\Rightarrow {S_n} = \sum {{n^2}} - \sum n + \sum 3
Now, we know that multiplication is repeated addition. So, 33 added n times is equal to 3n3n . So, we get,
⇒Sn=∑n2−∑n+3n\Rightarrow {S_n} = \sum {{n^2}} - \sum n + 3n
Now, we know the formula for the sum of the first n natural numbers is n(n+1)2\dfrac{{n\left( {n + 1} \right)}}{2}. So, substituting in the expression, we get,
⇒Sn=∑n2−n(n+1)2+3n\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 n(n+1)(2n+1)6\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}. Substituting this into the expression, we get,
⇒Sn=n(n+1)(2n+1)6−n(n+1)2+3n\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,
⇒Sn=n[(n+1)(2n+1)6−(n+1)2+3]\Rightarrow {S_n} = n\left[ {\dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \dfrac{{\left( {n + 1} \right)}}{2} + 3} \right]
⇒Sn=n[2n2+3n+16−(n+1)2+3]\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.
⇒Sn=n[2n2+3n+1−3n−3+186]\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,
⇒Sn=n[2n2+1−3+186]\Rightarrow {S_n} = n\left[ {\dfrac{{2{n^2} + 1 - 3 + 18}}{6}} \right]
⇒Sn=n[2n2+166]\Rightarrow {S_n} = n\left[ {\dfrac{{2{n^2} + 16}}{6}} \right]
Cancelling common factor in numerator and denominator, we get,
⇒Sn=[n3+8n3]\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.