Is there an easier way to prove this induction?

Question

Given that $u_1=1$, $u_{r+1} = \frac{2u_r-1}{3}$
Prove using induction that $u_n = 3(\frac{2}{3})^n-1$

Step 1: prove that $u_1=3(\frac{2}{3})^1-1$

1. $3(\frac{2}{3})^1-1$
2. $3(\frac{2}{3}) - 1$
3. $2-1$
4. $1$

Step 2: assume that $u_n = 3(\frac{2}{3})^n-1$

Step 3: prove that $u_{n+1} = 3(\frac{2}{3})^{n+1}-1$

1. $\frac{2[3(\frac{2}{3})^n-1]-1}{3} = 3(\frac{2}{3})^{n+1}-1$
2. $\frac{6(\frac{2}{3})^n-2-1}{3} = 3(\frac{2}{3})^{n+1}-1$
3. $2(\frac{2}{3})^n-1 = 3(\frac{2}{3})(\frac{2}{3})^{n}-1$
4. $2(\frac{2}{3})^n-1 = 2(\frac{2}{3})^{n}-1$

I was wondering if there is an easier way to prove it using induction.

Answer 1

You can simplify the argument for the inductive step: $$ \frac{2[3(\frac{2}{3})^n-1]-1}{3} = 3(\frac{2}{3})^{n+1} - \frac{2}{3} - \frac{1}{3} = 3(\frac{2}{3})^{n+1} - 1 $$ as desired.

If you were asking if there was an easier way to prove the relationship, you can solve the recurrence directly and then plug in the initial conditions...