Using induction to prove $\sum\limits^n_{k=1} 9^k = 0.5 \cdot \sum\limits^{2n}_{k=1} (-1)^k \cdot 3^{k+1}$

Question

$$\sum^n_{k=1} 9^k = 0.5 \cdot \left[\sum^{2n}_{k=1} (-1)^k \cdot 3^{k+1}\right]$$

I have tested both with a python script and it seems to be correct.

For the life of me, I am unable to unwind the $2n$ in the second sum in order to reconstruct the rhs sum to look like the lhs.

I am trying to practice with sigma notation - but this has been extremely frustrating so far. Any help would appreciated.

Edit: it is clear to me that I lack an understanding of Sigma Sum notation - or so I assume. What do you guys think? could you suggest resources I could use to better my understanding ?

Answer 1

Let $$\sum_{k=1}^n9^k=\frac12\left(\sum_{k=1}^{2n}(-1)^k3^{k+1}\right)$$

holds true for $n=m$

$$\implies\sum_{k=1}^m9^k=\frac12\left(\sum_{k=1}^{2m}(-1)^k3^{k+1}\right)$$

For $n=m+1$ $$\sum_{k=1}^{2(m+1)}(-1)^k3^{k+1}=\sum_{k=1}^{2m}(-1)^k3^{k+1}+(-1)^{2m+1}3^{2m+1+1}+(-1)^{2m+2}3^{2m+2+1}$$

$$=2\sum_{k=1}^m9^k-(-1)^{2k+1}3^{2m+2}(3-1)=2\sum_{k=1}^m9^k+2\cdot9^{m+1}$$

$$\implies \sum_{k=1}^{2(m+1)}(-1)^k3^{k+1}=2\sum_{k=1}^{m+1}9^k$$

Now establish the base case $n=1$