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