Question

Sum the following series: $x+a,{{x}^{2}}+2a,{{x}^{3}}+3a...$ to n terms.

Answer 1

Hint: Calculate the ${{n}^{th}}$ term of the series and then observe that each term of this series is a sum of terms of AP and terms of a GP. Calculate the sum of n terms of this AP and the sum of n terms of this GP and the two sums to get the value of n terms of the given series.

Complete step-by-step answer:
We have a series $x+a,{{x}^{2}}+2a,{{x}^{3}}+3a...$. We have to find the sum of n terms of this series.
We observe that the ${{n}^{th}}$ term of this series is $na+{{x}^{n}}$.
So, we have to find the value of $x+a+{{x}^{2}}+2a+{{x}^{3}}+3a...+na+{{x}^{n}}$.
We observe that each term of this series is written as a sum of terms of the AP $a,2a,3a,...$ and GP $x,{{x}^{2}},{{x}^{3}},...$, which means that the ${{n}^{th}}$ term of the given series is written as a sum of ${{n}^{th}}$ term of the GP and ${{n}^{th}}$ term of the AP.
So, to find the sum of the given series, we will find the sum of n terms of AP and n terms of GP and then add the two values to get the sum of the given series.
We have the AP $a,2a,3a,...$. We have to find the sum of first n terms of this AP. We observe that the first term of this AP is a and the common difference is $d=2a-a=a$.

We know that the sum of n terms of AP whose first term is ‘a’ and the common difference is ‘d’ is $\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.

Substituting $d=a$ in the above formula, we have $a+2a+3a+...na=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)a \right]=\dfrac{n}{2}\left[ 2a+na-a \right]=\dfrac{n}{2}\left[ a+na \right]=\dfrac{n}{2}\left( n+1 \right)a.....\left( 1 \right)$.

We will now calculate the sum of n terms of the GP $x,{{x}^{2}},{{x}^{3}},...$. We observe that the first term of GP is ‘x’ and the common ratio is $r=\dfrac{{{x}^{2}}}{x}=x$.
We know that sum of n terms of GP whose first term is ‘a’ and common ratio is ‘r’ is $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$.

Substituting $a=x,r=x$ in the above formula, we have $x+{{x}^{2}}+{{x}^{3}}+...{{x}^{n}}=\dfrac{x\left( {{x}^{n}}-1 \right)}{x-1}......\left( 2 \right)$.
We can rewrite $x+a+{{x}^{2}}+2a+{{x}^{3}}+3a...+na+{{x}^{n}}$ as $x+a+{{x}^{2}}+2a+{{x}^{3}}+3a...+na+{{x}^{n}}=\left( a+2a+...na \right)+\left( x+{{x}^{2}}+...+{{x}^{n}} \right)$.

Using equation (1) and (2), we have $x+a+{{x}^{2}}+2a+{{x}^{3}}+3a...+na+{{x}^{n}}=\dfrac{n}{2}\left( n+1 \right)a+\dfrac{x\left( {{x}^{n}}-1 \right)}{x-1}$.

Hence, the sum of n terms of the given series is $\dfrac{n}{2}\left( n+1 \right)a+\dfrac{x\left( {{x}^{n}}-1 \right)}{x-1}$.

Note: We must clearly know about any AP and GP. Arithmetic Progression (AP) is the sequence of numbers such that the difference between two consecutive terms is a constant. Geometric Progression (GP) is a sequence of numbers in which the ratio of two consecutive numbers is a constant.