# Sum the following series: \[x+a,{{x}^{2}}+2a,{{x}^{3}}+3a...\] to n terms.

Last updated date: 16th Mar 2023

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Answer

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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.

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.

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