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-
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 $\displaystyle \sum_{i = 1}^n{3^i} = \frac{3(3^n - 1)}{2}$.
Proof:
Base Step:
Let $\displaystyle n = 1$.
$\displaystyle LHS = 3^1 = 3$
$\displaystyle RHS = \frac{3(3^1 - 1)}{2} = \frac{3(2)}{2} = 3 = LHS.$
Inductive Step:
Assume the statement is true for $\displaystyle n = k$.
Then $\displaystyle \sum_{i = 1}^k{3^i} = \frac{3(3^k - 1)}{2}$.
Now let $\displaystyle n = k+1$.
Then we wish to show $\displaystyle \sum_{i = 1}^{k + 1}{3^i} = \frac{3(3^{k + 1} - 1)}{2}$.
$\displaystyle LHS = \sum_{i = 1}^{k + 1}{3^i}$
$\displaystyle = \sum_{i = 1}^{k}{3^i} + 3^{k + 1} $
$\displaystyle = \frac{3(3^k - 1)}{2} + 3^{k + 1}$
$\displaystyle = \frac{3(3^k - 1) + 2(3^{k + 1})}{2}$
$\displaystyle = \frac{3^{k + 1} - 3 + 2(3^{k + 1})}{2}$
$\displaystyle = \frac{3(3^{k + 1}) - 3}{2} $
$\displaystyle = \frac{3(3^{k + 1} - 1)}{2}$
$\displaystyle = RHS$.
QED.
Hello, AlgebraicallyChallenged!
Are you sure you understand the Induction Procedure?
We want to prove .$\displaystyle S(n):\;\;3 + 3^2 + 3^3 + \hdots + 3^n \;=\;\frac{3(3^n-1)}{2}$Prove by mathematical induction: .$\displaystyle \sum^n_{i=1} 3^i \;=\;\frac{3(3^n-1)}{2}$
Verify $\displaystyle S(1)\!:\;\;3^1 \:=\:\frac{3(3-1)}{2} \:=\:3$ . . . True!
Assume $\displaystyle S(k)\!:\;\;3 + 3^2 + 3^3 + \hdots + 3^k \;=\;\frac{3(3^k-1)}{2}$ .[1]
Now we must prove $\displaystyle S(k+1)\!:\;\;3 + 3^2 + 3^3 + \hdots + 3^{k+1} \;=\;\frac{3(3^{k+1}-1)}{2}$ .[2]
Start with [1]: .$\displaystyle 3 + 3^2 + 3^3 + \hdots + 3^k \;=\;\frac{3(3^k-1)}{2}$
$\displaystyle \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: .$\displaystyle \frac{3(3^k-1)}{2} + 3^{k+1} \;\;=\;\;\frac{3^{k+1}}{2} - \frac{3}{2} + 3^{k+1} $
. . $\displaystyle = \;\;\frac{1}{2}3^{k+1} + 3^{k+1} - \frac{3}{2} \;\;=\;\;\frac{3}{2}3^{k+1} - \frac{3}{2}$
. . $\displaystyle =\;\;\frac{3}{2}\left(3^{k+1}-1\right) \;\;=\;\;\underbrace{\frac{3(3^{k+1}-1)}{2}}_{\text{right side of }{\color{blue}[2]}}$
We have proved $\displaystyle S(k+1)$ . . .The inductive proof is compete.