1. ## [SOLVED] Induction Problems

Dear Forum I am having major troubles with the following question I tried to set both sides equal to one another and still can't figure it out. If anyone can help me tackle this confusing induction issue it would be greatly appreciated.

Thanks -AC-

2. Originally Posted by AlgebraicallyChallenged
Dear Forum I am having major troubles with the following question I tried to set both sides equal to one another and still can't figure it out. If anyone can help me tackle this confusing induction issue it would be greatly appreciated.

Thanks -AC-
Prove $\sum_{i = 1}^n{3^i} = \frac{3(3^n - 1)}{2}$.

Proof:

Base Step:

Let $n = 1$.

$LHS = 3^1 = 3$

$RHS = \frac{3(3^1 - 1)}{2} = \frac{3(2)}{2} = 3 = LHS.$

Inductive Step:

Assume the statement is true for $n = k$.

Then $\sum_{i = 1}^k{3^i} = \frac{3(3^k - 1)}{2}$.

Now let $n = k+1$.

Then we wish to show $\sum_{i = 1}^{k + 1}{3^i} = \frac{3(3^{k + 1} - 1)}{2}$.

$LHS = \sum_{i = 1}^{k + 1}{3^i}$

$= \sum_{i = 1}^{k}{3^i} + 3^{k + 1}$

$= \frac{3(3^k - 1)}{2} + 3^{k + 1}$

$= \frac{3(3^k - 1) + 2(3^{k + 1})}{2}$

$= \frac{3^{k + 1} - 3 + 2(3^{k + 1})}{2}$

$= \frac{3(3^{k + 1}) - 3}{2}$

$= \frac{3(3^{k + 1} - 1)}{2}$

$= RHS$.

QED.

3. Hello, AlgebraicallyChallenged!

Are you sure you understand the Induction Procedure?

Prove by mathematical induction: . $\sum^n_{i=1} 3^i \;=\;\frac{3(3^n-1)}{2}$
We want to prove . $S(n):\;\;3 + 3^2 + 3^3 + \hdots + 3^n \;=\;\frac{3(3^n-1)}{2}$

Verify $S(1)\!:\;\;3^1 \:=\:\frac{3(3-1)}{2} \:=\:3$ . . . True!

Assume $S(k)\!:\;\;3 + 3^2 + 3^3 + \hdots + 3^k \;=\;\frac{3(3^k-1)}{2}$ .[1]

Now we must prove $S(k+1)\!:\;\;3 + 3^2 + 3^3 + \hdots + 3^{k+1} \;=\;\frac{3(3^{k+1}-1)}{2}$ .[2]

Start with [1]: . $3 + 3^2 + 3^3 + \hdots + 3^k \;=\;\frac{3(3^k-1)}{2}$

$\text{Add }3^{k+1}\text{ to both sides: }\;\underbrace{3 + 3^2 + 3^3 + \hdots + 3^{k+1}}_{\text{left side of }{\color{blue}[2]}} \;=\;\frac{3(3^k-1)}{2} + 3^{k+1}$

The left side is already the left side of [2].
We must show that the right side is the right side of [2].

On the right, we have: . $\frac{3(3^k-1)}{2} + 3^{k+1} \;\;=\;\;\frac{3^{k+1}}{2} - \frac{3}{2} + 3^{k+1}$

. . $= \;\;\frac{1}{2}3^{k+1} + 3^{k+1} - \frac{3}{2} \;\;=\;\;\frac{3}{2}3^{k+1} - \frac{3}{2}$

. . $=\;\;\frac{3}{2}\left(3^{k+1}-1\right) \;\;=\;\;\underbrace{\frac{3(3^{k+1}-1)}{2}}_{\text{right side of }{\color{blue}[2]}}$

We have proved $S(k+1)$ . . .The inductive proof is compete.