Question

How to calculate this sum: $S = 1 + 2x + 3{x^2} + 4{x^3} + .....n{x^{\left( {n - 1} \right)}}$ ?

Answer 1

Hint: In the question, we are required to find the sum of an arithmetic geometric progression. So, we follow a structured method to find the same. Arithmetic geometric progression is a series whose terms involve the product of two types of terms. One term is in arithmetic progression and the other one is in geometric progression.

Complete step by step solution:
So, we need to find the sum of the given arithmetic geometric progression.
So, $S = 1 + 2x + 3{x^2} + 4{x^3} + .....n{x^{\left( {n - 1} \right)}}$
In this arithmetic geometric progression, the x component is in geometric progression and the constant term is arithmetic progression.
Hence, we multiply both sides of the equation by the common ratio of the geometric progressive terms, x.
So, $S = 1 + 2x + 3{x^2} + 4{x^3} + .....n{x^{\left( {n - 1} \right)}} - - - - - - - - (1)$
$ \Rightarrow xS = x\left[ {1 + 2x + 3{x^2} + 4{x^3} + .....n{x^{\left( {n - 1} \right)}}} \right] $
Opening the bracket and simplifying, we get,
$ \Rightarrow xS = x + 2{x^2} + 3{x^3} + 4{x^4} + ......n{x^n} - - - - - - (2)$
Now, subtracting $(2)$from $(1)$, we get,
$ \Rightarrow \left( {1 - x} \right)S = \left[ {1 + x + {x^2} + {x^3}.....{x^{\left( {n - 1} \right)}}} \right] - n{x^n}$
The term contained in square brackets is a geometric progression. Hence, we can find the sum of n terms of a geometric progression by formula: $a\dfrac{{\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$
 \[ \Rightarrow \left( {1 - x} \right)S = \left( {1\dfrac{{\left( {1 - {x^n}} \right)}}{{\left( {1 - x} \right)}}} \right) - n{x^n}\]
 \[ \Rightarrow \left( {1 - x} \right)S = \left( {\dfrac{{1 - {x^n} - n{x^n} + n{x^{\left( {n + 1} \right)}}}}{{1 - x}}} \right)\]
 \[ \Rightarrow S = \left( {\dfrac{{1 - {x^n} - n{x^n} + n{x^{\left( {n + 1} \right)}}}}{{{{\left( {1 - x} \right)}^2}}}} \right)\]
Hence the sum of the given arithmetic geometric progression: $S = 1 + 2x + 3{x^2} + 4{x^3} + .....n{x^{\left( {n - 1} \right)}}$ is \[S = \left( {\dfrac{{1 - {x^n} - n{x^n} + n{x^{\left( {n + 1} \right)}}}}{{{{\left( {1 - x} \right)}^2}}}} \right)\] .
So, the correct answer is “ \[S = \left( {\dfrac{{1 - {x^n} - n{x^n} + n{x^{\left( {n + 1} \right)}}}}{{{{\left( {1 - x} \right)}^2}}}} \right)\] ”.

Note: The method involved in the problem given to us must be remembered as it is a standard method that can be applied in various questions involving arithmetic geometric progressions or any other type of special series or progression. We also must remember the formula for finding the sum of n terms of a geometric progression: $a\dfrac{{\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$ and the formula for finding the sum of an infinite geometric progression $\dfrac{a}{{1 - r}}$.