# Find the sum of the following A.G.P

Find the sum of: $$1×2+ 2×3x+ 3×4x^2...$$ I tried the problem and I am getting answer as $$\frac{(2-x)}{(1-x)^2}$$ which I think is wrong Can Someone please tell the correct answer so that I can find my mistake

• Why don’t you show your calculations so that we can point out the error? – Martin R May 13 at 6:59

Use the following. $$1\cdot2+2\cdot3x+3\cdot4x^2+...=\left(2x+3x^2+4x^3+...\right)'=(x^2+x^3+x^4+...)''$$ For $$-1 I got $$\frac{2}{(1-x)^3}.$$

$$S = 1×2+ 2×3x+ 3×4x^2+4×5x^3...$$

$$Sx = ~~~~~~~~~~~1×2x+ 2×3x^2+ 3×4x^3...$$(multiplying $$S$$ with $$x$$)

Subtracting,

$$(S-Sx) = 2+4x+6x^2+8x^3+\dots\Rightarrow 2(1+2x+3x^2+4x^3+\dots) = 2L$$(suppose L is that sum.)

$$L = 1+ 2x+ 3x^2+4x^3+\dots$$

$$Lx = ~~~~~~x+2x^2+3x^3+\dots$$

Subtracting,

$$L(1-x) = \frac{1}{1-x} \Rightarrow L= \frac{1}{(1-x)^2}$$

Putting the value of $$L$$ in the above equation, $$S(1-x)=2L=2\times\frac{1}{(1-x)^2}$$, therefore, $$\boxed{S= \frac{2}{(1-x)^3}}$$

The sum is: $$\sum_{i=1}^{\infty} i(i+1)x^{i-1}$$

The standard method is: \begin{align}f(x)&=\sum_{i=0}^{\infty}x^{i+1}=\frac{x}{1-x},|x|<1 \Rightarrow \\ f'(x)&=\sum_{i=0}^{\infty} (i+1)x^i=\frac{1}{(1-x)^2} \Rightarrow \\ f''(x)&=\sum_{i=1}^\infty i(i+1)x^{i-1}=\frac{2}{(1-x)^3}.\end{align}