Question

Find the sum of n terms of the series 3 + 15 + 35 + 63 + _ _.

Answer 1

Hint: Let us find the \[{n^{th}}\] term of the above given series and then apply summation to find the sum of n terms of that series.

Complete Step-by-Step solution:
Now we had to reduce the above given series such that its \[{n^{th}}\] term can be found easily.
Let S be the sum of n terms of the above given series.
And let the \[{n^{th}}\] term of the given series be denoted as \[{T_n}\].
S = 3 + 15 + 35 + 63 + …… + \[{T_n}\] (1)
Now, if we add 1 to each term of the above series of equation 1. Then equation 1 becomes,
S + n = (3 + 1) + (15 + 1) + (35 + 1) + (63 + 1) + …….. + (\[{T_n}\] + 1)
On solving above equation, it becomes
S + n = 4 + 16 + 36 + 64 + ……………… (\[{T_n}\] + 1)
Now, let \[{n^{th}}\] term of the above series is \[{P_n}\].
S + n = \[{\left( 2 \right)^2} + {\left( 4 \right)^2} + {\left( 6 \right)^2} + {\left( 8 \right)^2} + ........\left( {{P_n}} \right)\] (2)
So, we can clearly see from the above equation that,
\[{n^{th}}\] term of the series at equation 2 will be equal to \[{P_n}\] = \[{\left( {2n} \right)^2}\].
Like, if n = 1 then the first term will be = \[{\left( {2 \times 1} \right)^2}\] = 2
And if n = 2 then the second term will be = \[{\left( {2 \times 2} \right)^2}\] = 16
Let the sum of n terms of RHS of equation 2 is denoted as \[S{P_n}\].
S + n = \[S{P_n}\] (3)
So, now applying summation from i = 1 to i = n to the \[{i^{th}}\] term of the series whose sum of n terms is \[S{P_n}\] we get.
\[S{P_n} = \sum\limits_{i = 1}^{i = n} {{{(2i)}^2} = } 4\sum\limits_{i = 1}^{i = n} {{{(i)}^2}} = 4\left[ {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 5 \right)}^2} + .......... + {{\left( n \right)}^2}} \right]\] (4)
Now as we know that if the \[{i^{th}}\] term of the series is \[{(i)^2}\] then the sum of n terms of that series will be equal to,
\[{\left( 1 \right)^2} + {\left( 2 \right)^2} + {\left( 3 \right)^2} + {\left( 4 \right)^2} + {\left( 5 \right)^2} + .......... + {\left( n \right)^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]
So, equation 4 becomes
\[S{P_n}\] = \[\dfrac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]
Now putting the value of \[S{P_n}\] in equation 3. We get,
S + n = \[\dfrac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]
Now subtracting n to both sides of the above equation. We get,
S = \[\dfrac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\] – n
 Now taking LCM on RHS of the above equation. We get,
S = \[\dfrac{{4n\left( {n + 1} \right)\left( {2n + 1} \right) - 6n}}{6} = \dfrac{{n\left[ {4\left( {n + 1} \right)\left( {2n + 1} \right) - 6} \right]}}{6}\]
On solving the above equation. We get,
S = \[\dfrac{{n\left[ {8{n^2} + 4n + 8n + 4 - 6} \right]}}{6} = \dfrac{{n\left[ {8{n^2} + 12n - 2} \right]}}{6} = \dfrac{{8{n^3} + 12{n^2} - 2n}}{6} = \dfrac{{4{n^3} + 6{n^2} - 1n}}{3}\]
As we have stated above that S is the sum of n terms of the series given in the question.
So, the sum of n terms of the given series will be equal to \[\dfrac{{4{n^3} + 6{n^2} - 1n}}{3}\].

Note: Whenever we come up with this type of question then if the given series does not form any relation like A.P or G.P directly. Then to find the sum of n terms of that series first we try to reduce that series by adding or subtracting some constant term to each term of that series after that we had to find the \[{n^{th}}\] term of the new series and then apply summation from i = 1 to i = n to get the sum of n terms of the new series. After that we add or subtract n times the constant term which we had added or subtracted earlier to get the required sum of n terms of the given series.