I was recently sent the following question via PM (which we discourage here), so I thought I would start a thread and help here (and give the OP a link to this thread). Here's the question:

(a) We can observe that we may write:

$ \displaystyle 2\cdot\sum_{k=0}^{n}\left(3^k\right)=2\cdot\sum_{k=0}^{n-1}\left(3^k\right)+2\cdot3^n$

Hence:

$ \displaystyle S(n)=S(n-1)+2\cdot3^n$

Now, if we were to derive a closed form for $S(n)$ from the above linear inhomogeneous recursion, we would observe that the homogeneous solution is:

$ \displaystyle h_n=c_1$

And, we would be looking for a particular solution of the form:

$ \displaystyle p_n=A3^n$

So, to use the method of undetermined coefficients, we would substitute $p_n$ into the recursion to get:

$ \displaystyle A3^n-A3^{n-1}=2\cdot3^n$

Divide through by $3^{n-1}$:

$ \displaystyle 3A-A=6\implies A=3$

Thus:

$ \displaystyle p_n=3^{n+1}$

Hence, by the principle of superposition, we find the general solution to be:

$ \displaystyle S(n)=h_n+p_n=3^{n+1}+c_1$

Now, to determine the parameter $c_1$, we find:

$ \displaystyle S(1)=2(1+3)=8=3^{1+1}+c_1=9+c_1\implies c_1=-1$

And so we have:

$ \displaystyle S(n)=3^{n+1}-1$

This agrees with the induction hypothesis we are to prove for part (b).

(b) State the induction hypothesis $ \displaystyle P_n$:

$ \displaystyle 2\cdot\sum_{k=0}^{n}\left(3^k\right)=3^{n+1}-1$

Show the base case $P_1$ is true:

$ \displaystyle 2\cdot\sum_{k=0}^{1}\left(3^k\right)=3^{1+1}-1$

$ \displaystyle 2(3^0+3^1)=3^{1+1}-1$

$ \displaystyle 8=8\quad\checkmark$

The base case is true, so as our induction step, we may add $2\cdot3^{n+1}$ to each side of $P_n$:

$ \displaystyle 2\cdot\sum_{k=0}^{n}\left(3^k\right)+2\cdot3^{n+1}=3^{n+1}-1+2\cdot3^{n+1}$

One the left, incorporate the added term within the sum while on the right combine like terms:

$ \displaystyle 2\cdot\sum_{k=0}^{n+1}\left(3^k\right)=3\cdot3^{n+1}-1$

$ \displaystyle 2\cdot\sum_{k=0}^{n+1}\left(3^k\right)=3^{(n+1)+1}-1$

We have derived $P_{n+1}$ from $P_n$, thereby completing the proof by induction.