find the sum of 1.2+2.3x+3.4x² +....... (till infinity) (mod of x < 1)

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$$\begin{gathered} \mathrm{By} \mathrm{binomial} \mathrm{theorem} \mathrm{of} \mathrm{any} \mathrm{index}, \mathrm{we} \mathrm{have}: \\ {\left(1-\mathrm{x}\right)}^{-\mathrm{n}}=1+\mathrm{nx}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2!}{\mathrm{x}}^2+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}{3!}{\mathrm{x}}^3+...... \\ \mathrm{In} \mathrm{required} \mathrm{series}, \mathrm{first} \mathrm{term} \mathrm{is} 2. \mathrm{Hence} \mathrm{multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} 2, \mathrm{we} \mathrm{get}: \\ 2{\left(1-\mathrm{x}\right)}^{-\mathrm{n}}=2+2\mathrm{nx}+2\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2!}{\mathrm{x}}^2+2\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}{3!}{\mathrm{x}}^3+...... \\ \mathrm{Now} \mathrm{second} \mathrm{term} \mathrm{is} 2.3.\mathrm{x}, \mathrm{hence} \mathrm{put} \mathrm{n}=3 \\ \Rightarrow 2{\left(1-\mathrm{x}\right)}^{-3}=2+2.3.\mathrm{x}+2\frac{3\left(3+1\right)}{2!}{\mathrm{x}}^2+2\frac{3\left(3+1\right)\left(3+2\right)}{3!}{\mathrm{x}}^3+...... \\ \Rightarrow \frac{2}{{\left(1-\mathrm{x}\right)}^3}=1.2+2.3.\mathrm{x}+2\frac{3.4}{2}{\mathrm{x}}^2+2\frac{3.4.5}{3.2.1}{\mathrm{x}}^3+...... \\ \Rightarrow \frac{2}{{\left(1-\mathrm{x}\right)}^3}=1.2+2.3.\mathrm{x}+3.4.{\mathrm{x}}^2+4.5.{\mathrm{x}}^3+.... \\ \mathrm{i}.\mathrm{e}. \\ \mathbf{1}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{.}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{.}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{.}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{=}\frac{\mathbf{2}}{{\left(\mathbf{1}\mathbf{-}\mathbf{x}\right)}^{\mathbf{3}}}\mathbf{ }\left[\mathbf{Answer}\right] \end{gathered}$$

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