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?

Why did you write it as
$9(3^{k-1} -1/ 2) + 3^{k+1}  = 9((3^k - 1)/2)$
Rather than $9((3^{k-1} -1)/ 2) + 3^{k+1}  = 9((3^k - 1)/2)$?
Anyway
$9\frac {3^{k-1} - 1}2 + 3^{k+1}=$
$9\frac {3^{k-1} - 1}2 + \frac {2*3^{k+1}}2=$
$9\frac {3^{k-1} - 1}2 + \frac {2*9*3^{k-1}}2 = $
$9\frac {3^{k-1} -1 + 2*3^{k-1}}2=$
$9\frac {3*3^{k-1} -1}2 =$
$9\frac {3^k -1}2$
