Sum the following series: \[x+a,{{x}^{2}}+2a,{{x}^{3}}+3a...\] to n terms.
Answer
Verified358.8k+ views
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.
Last updated date: 19th Sep 2023
•
Total views: 358.8k
•
Views today: 5.58k
