Trouble proving $3^2 + 3^3 + ... 3^n = 9 \cdot \frac{3^{n-1} - 1}2$ by induction So i'm supposed to prove by mathematical induction that this formula: $3^2 + 3^3 + ... 3^n = 9 \cdot \dfrac{3^{n-1} - 1}2$ holds true for all numbers greater than 2.
I started with the base case and just plugged in 2, it worked.
Then I assumed k was true, and then considered k+1.
What I ended up with is:
$9(3^{k-1} -1/ 2) + 3^{k+1}  = 9((3^k - 1)/2)$
I tried making the left handside equal to the right handside but I couldn't. Did I do something wrong along the way?

Suppose $t=3^{n-1}$
Then:
$$\frac 92 (t-1)+9t=\frac 92t+9t-\frac92=\frac{27}{2}t-\frac92$$
$$=\frac92(3t-1)$$
Then sub back in, we get: $\frac 92(3^n-1)$ as required.
