Question

Find the sum of the series $ 1 + 2x + 3{x^2} + 4{x^3} + ... $ if $ |x| < 1 $
a) $ \dfrac{{ - 1}}{{{{(1 - x)}^2}}} $
b) $ \dfrac{1}{{{{(1 - x)}^2}}} $
c) $ \dfrac{1}{{{{(1 + x)}^2}}} $
d) $ \dfrac{{ - 1}}{{{{(1 + x)}^2}}} $

Answer 1

Hint: We will use the series $ 1 + x + {x^2} + {x^3} + ... $ to evaluate the problem. This series has a unique property when $ |x| < 1 $ and we are going to use that to get the result. We will simplify the series $ 1 + 2x + 3{x^2} + 4{x^3} + ... $ in terms of the series $ 1 + x + {x^2} + {x^3} + ... $ by algebraic manipulations which will enable us to find the exact expression for this infinite series.

Complete step-by-step answer:
If $ |x| < 1 $ then the series,
 $ 1 + x + {x^2} + {x^3} + ... = \dfrac{1}{{1 - x}} $
The given infinite series is:
 $ 1 + 2x + 3{x^2} + 4{x^3} + ... $ with the condition that $ |x| < 1 $
Let,
 $ S = 1 + 2x + 3{x^2} + 4{x^3} + ... $ --(1)
Let’s multiply $ x $ on both sides of the equation to get a new form of the series
 $ xS = x(1 + 2x + 3{x^2} + 4{x^3} + ...) $
Simplifying the above we get:
 $ Sx = x + 2{x^2} + 3{x^3} + 4{x^4} + ... $ --(2)
Subtracting (2) from (1) we get:
 $ Sx - S = (x + 2{x^2} + 3{x^3} + 4{x^4} + ...) - (1 + 2x + 3{x^2} + ...) $
Keep the like terms separate so that we can simplify the above equation to the simplest form as:
 $ S(x - 1) = ( - 1) + (x - 2x) + (2{x^2} - 3{x^2}) + (3{x^3} - 4{x^3}) + ... $
Subtracting the separated terms inside brackets we get the following:
 $ S(x - 1) = ( - 1) + ( - x) + ( - {x^2}) + ( - {x^3}) + ... $
 $ \Rightarrow S(x - 1) = - (1 + x + {x^2} + {x^3} + ...) $
 $ \Rightarrow - S(1 - x) = - (1 + x + {x^2} + {x^3} + ...) $
 $ \Rightarrow S(1 - x) = 1 + x + {x^2} + {x^3} + ... $ [negative sign on both sides cancelled with each other]
The expression in the R.H.S is same as $ \dfrac{1}{{1 - x}} $ , so we can write it as:
 $ S(1 - x) = \dfrac{1}{{1 - x}} $
On cross multiplication we get:
 $ S = \dfrac{1}{{{{(1 - x)}^2}}} $
 $ \Rightarrow 1 + 2x + 3{x^2} + 4{x^3} + ... = \dfrac{1}{{{{(1 - x)}^2}}} $
Thus, the sum of the series $ 1 + 2x + 3{x^2} + 4{x^3} + ... $ if $ |x| < 1 $ is $ \dfrac{1}{{{{(1 - x)}^2}}} $ .
The option that corresponds to our calculated value is ‘b’ so it is the correct option.
So, the correct answer is “Option D”.

Note: The series equality holds only if the condition $ |x| < 1 $ is satisfied. You can check it by taking different values of $ x $ like $ x = 1,x = 2,x = 0.5 $ to get a clear picture of why the condition $ |x| < 1 $ should be satisfied. Keeping this in mind don’t forget to mention this condition after solving the problem because this is the most important and sensitive thing in this problem.