Trouble proving $3^2 + 3^3 + ... 3^n = 9 \cdot \frac{3^{n-1} - 1}2$ by induction

Question

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?

Answer 1

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.

Answer 2

Left side is $\frac {3^{k+1}} 2-\frac 9 2 +3^{k+1}=3^{k+1} (\frac 3 2)-\frac 9 2$ and right side is same: just write $9$ as $3^{2}$ on both sides.

Answer 3

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 = $

$9\frac {3^{k-1} -1 + 2*3^{k-1}}2=$

$9\frac {3*3^{k-1} -1}2 =$

$9\frac {3^k -1}2$