What is the sum from n=1 to infinity}(2^nx^{2n of)/(3^{n+1)} ?

The sum from n=1 to infinity}(2^nx^{2n of)/(3^{n+1)} is Radius of convergence is sqrt(3/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=1 to infinity}(2^nx^{2n of)/(3^{n+1)}

n = 1 \sum \infty \frac{2 ^{n} x ^{2 n}}{3 ^{n + 1}}

Radius of convergence is \frac{3}{2}

Solution steps

n = 1 \sum \infty \frac{2 ^{n} x ^{2 n}}{3 ^{n + 1}}

Use the Ratio Test to compute the convergence interval

Radius of convergence is \frac{3}{2}

\frac{\partial}{\partial x} (4 arctan (\frac{3}{x}))

f (x) = 4 + 5 x^{2}, (0, 4)

y^{^{''}} + 2 y^{^{'}} + 5 y = 8 sin (2 t)

\frac{\partial}{\partial x} (y cos (x) + 2 x e^{y})

\int_{0}^{25} \frac{1}{25 - x} d x

What is the sum from n=1 to infinity}(2^nx^{2n of)/(3^{n+1)} ?
The sum from n=1 to infinity}(2^nx^{2n of)/(3^{n+1)} is Radius of convergence is sqrt(3/2)