calculus - Solving the infinite sum $sum_{k=0}^{infty} left(frac{1}{3}right)^{k+3}$

I'm stuck on the question $\sum_{k=0}^{\infty} \left(\frac{1}{3}\right)^{k+3}$

I know that $\sum_{k=0}^{\infty} \left(\frac{1}{3}\right)^k$ is solved by using $\sum_{k=0}^{\infty} a^{k} =$ $\frac{1}{1-a}$ and the answer is $\frac{3}{2}$

So is there a way I could apply that to the above question or is there a different way to approach the problem?

Answer

$$\sum_{k=0}^n\left(\frac{1}{3}\right)^{k+3}=\frac{1}{3^3}\sum_{k=0}^n\left(\frac{1}{3}\right)^{k}\to\frac{1}{27}\frac{1}{1-\frac{1}{3}}=\frac{1}{27}\frac{3}{2}=\frac{1}{18}$$