What is the sum from n=3 to infinity of 4^{(n+2)} ?

The sum from n=3 to infinity of 4^{(n+2)} is diverges

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sum from n=3 to infinity of 4^{(n+2)}

n = 3 \sum \infty 4^{(n + 2)}

diverges

Solution steps

n = 3 \sum \infty 4^{n + 2}

Apply Series Ratio Test:

diverges

= diverges

\int \frac{x ^{2} + x + 1}{x} d x

(1 + x y) y^{^{'}} + y^{2} = 0

y = (arctan (5 x))^{2}

k = 0 \sum \infty (\frac{x - 4}{5})^{k}

\frac{\partial}{\partial x} (2 x sin (y - z))

What is the sum from n=3 to infinity of 4^{(n+2)} ?
The sum from n=3 to infinity of 4^{(n+2)} is diverges