Question

Find the sum of the series. The sum from n=0 to infinity of 2^(n+2)/3^n

Answer 1

\(\large \color{black}{

\sum_{i=0}^{\infty}\frac{2^{n+2}}{3^n}=\sum_{i=0}^{\infty}\frac{2^n\cdot 2^2}{3^n}\hspace{.33em}\\~\\

=4\sum_{i=0}^{\infty}\frac{2^n}{3^n}\hspace{.33em}\\~\\
\implies \sum_{i=0}^{\infty}\left(\frac{2}{3}\right)^n=\left(\frac{2}{3}\right)^0+\left(\frac{2}{3}\right)^1+\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^3+\cdots \infty \hspace{.33em}\\~\\
\normalsize \text{let }\sum_{i=0}^{\infty}\frac{2^{n}}{3^n}=x\hspace{.33em}\\~\\
x=\left(\frac{2}{3}\right)^0+\left(\frac{2}{3}\right)^1+\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^3+\ldots +\infty \hspace{.33em}\\~\\
x=1+\left(\frac{2}{3}\right)\left(1+\left(\frac{2}{3}\right)^1+\left(\frac{2}{3}\right)^2+\ldots \infty\right) \hspace{.33em}\\~\\
x=1+\left(\frac{2}{3}\right)\left(x\right) \hspace{.33em}\\~\\
3x=\left(3+2x\right) \hspace{.33em}\\~\\
x=3\hspace{.33em}\\~\\
4\sum_{i=0}^{\infty}\frac{2^{n}}{3^n}=4x=4\times 3=12\hspace{.33em}\\~\\
}\)