Find closed form of $\sum_{j=1}^n 3^{j+1}$

Question

What is the closed form of the following:

$$\sum_{j=1}^n 3^{j+1}$$

I'm new to summations. Is it this?

$$\sum_{j=1}^n 3^{j} + \sum_{j=1}^n 3$$

Then using the closed form formula:

$$\frac{3^{n+1} - 1}{2} + 3n$$

Answer 1

No. Just factor out $3^2$ so as to obtain the standard geometric series: $$ \sum_{j=1}^n 3^{j+1}=9 \sum_{j=0}^{n-1}3^j=9\frac{3^n-1}{3-1}=\frac{3^{n+2}-3^2}{2}. $$

Answer 2

No, $3^{j+1}=3^j\times 3$.

So multiply by $3$ the sum $$\sum_{j=1}^n 3^j.$$

Plus you got this summation wrong, because

$$\sum_{j=1}^n 3^j=\frac{3^n-1}{3-1}=\frac{3^n-1}{2}$$

instead of $n+1$.

The result is then:

$$\frac {3^{n+1}-3}2.$$

Answer 3

Hint: This is the finite geometric series.

$$S_0=3\sum_{j=1}^{n}3^j=3S_n$$

We will only look at $S_n$ from now on. $S_n=3+3^2+3^3+\dots+3^n$ so multiply with 3 to get $3S_n=3^2+3^3+\dots+3^{n+1}$. Subtract both equations and notice the chancellations to get $S_n-3S_n=3-3^{n+1}$. Solve for $S_n$ and resubstitute into $S_0=3S_n$ to get your closed solutions.