Find a sum of a series

Question

Help me find a sum of this series

I tried to excrete as (2/7)^n * 3^(n+2) and use De Lamber indication. It gives me a result 6/7.

I checked it in Wolfram Math but the result was 54. Where did I go wrong? Thanks

Answer 1

$$\sum_{n=1}^{\infty}\frac{2^n3^{n+2}}{7^n}=3^2\sum_{n=1}^{\infty}\left(\frac67\right)^n$$

Using Infinite Geometric Series formula, this equals to $$\frac{\left(\dfrac67\right)^1}{1-\dfrac67}=\cdots$$ as $\displaystyle\left|\frac67\right|<1$

Answer 2

Factor out $3^2$ and you get $\sum_{n=1}^\infty (6/7)^n$ which you should have a geometric series formula for (you may have to add 1 so that you get the sum starting at $n=0$ for your formula to work).