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

The sum of the series is _________.
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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.$x^i$ = $\frac{x}{(1 - x)^2}$

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