Test by Bikram | Mathematics | Test 2 | Question: 10

Question

$S = 1.2 + 2.3.x + 3.4.x^2 + 4.5.x^3 + \dots +\infty$ where $x = 0.5$.

The sum of the series is _________.

Comments

Can someone solve this?

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@Harish Karnam,

See the below answer.

Answer 1

S = 1.2 + 2.3.x + 3.4.$x^2$ + 4.5.$x^3$ +........$\infty$    Where x = 0.5

xS = 1.2.x + 2.3.$x^2$ + 3.4.$x^3$ + 4.5.$x^4$ + .........$\infty$.

Now,

S - xS = 1.2 + (2.3 - 1.2)x + (3.4 - 2.3)$x^2$ + (4.5 - 3.4)$x^3$ + (5.6 - 4.5)$x^4$ + ..........$\infty$

(1 - x)S = 1.2 + 2.2.x + 3.2.$x^2$ + 4.2.$x^3$ + 5.2.$x^4$ + ..........$\infty$

Taking 2 as common.

(1 - x)S = 2 * (1 + 2.x + 3.$x^2$ + 4.$x^3$ + 5.$x^4$ + .......$\infty$)

(1 - x)S = 2 * $\sum_{i=0}^{\infty }$ (i + 1).$x^i$

(1 - x)S = 2 * [ $\sum_{i=0}^{\infty }$ i.$x^i$  +  $\sum_{i=0}^{\infty }$ $x^i$ ]

(1 - x)S = 2 * [ $\frac{x}{(1 - x)^2}$  + $\frac{1}{(1 - x)}$ ]

Put x = 0.5

S/2 = 2 * [ 2 + 2]

S = 16. Ans.

Proof: $\sum_{i=0}^{\infty }$ i.$x^i$ = $\frac{x}{(1 - x)^2}$

We already know the sum of infinity G.P.

$\sum_{i=0}^{\infty }$ $x^i$ = $\frac{1}{(1 - x)}$

Differentiate w.r.t to x on both sides, we get.

$\sum_{i=0}^{\infty }$ i.$x^{i-1}$ = $\frac{1}{(1 - x)^2}$

Now multiply on both side with x.

$\sum_{i=0}^{\infty }$ i.$x^i$ = $\frac{x}{(1 - x)^2}$

Comments

(1 - x)S = 2 * (1 + 2.x + 3.x² + 4.x³ + 5.x⁴ + .......∞)-----------------------equation i

now, say, J=1 + 2.x + 3.x² + 4.x³ + 5.x⁴ + .......∞

               xJ=        x+ 2.x² + 3.x³ + 4.x⁴ + 5.x⁵ + .......∞

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            (1-x)J=1+x+x²+x³+..........................

                    J=$\frac{1-x^{n+1}}{\left ( 1-x \right )^{2}}$

Now, putting in equation i we get

$S=2\times \frac{1-x^{n+1}}{\left ( 1-x \right )^{3}}$

but here what should be value of n?

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@srestha, n is infinity and x is 0.5 (< 1). We can use the formula for the sum of infinite series.

$S_n$ = $\frac{a}{(1-r)}$

Answer 2

Multiply both sides by x and subtract from original equation. use infinite arithmetic geometric progression formula. 

Formula for Infinite Series:

https://en.wikipedia.org/wiki/Arithmetico–geometric_sequence#Infinite_series

Answer 3

 might be helpful