Prove by induction:$\sum_{i=0}^n 3^i = \frac {(3^{n+1})-1}{2}$

Question

Prove by induction: $$\sum_{i=0}^n 3^i = \frac {(3^{n+1})-1}{2}$$

Basis: For $n=0$ we have $1 = 1$

Inductive Step: Now this is where I don't know what to do, any kind of help would be much appreciated. Thanks!

Have you tried anything yourself? If you have any idea as to how the proof might go then indicating this would be useful. — Joe Tait, Apr 01 '14 at 08:44

score 1 · Accepted Answer · answered Apr 01 '14 at 08:27

Inductive Step: Assume that the expression holds for $n$. We want to prove that the expression holds also for $n+1$. That is $$\begin{align*}\sum_{i=0}^{n+1} 3^i &=\sum_{i=0}^n 3^i +3^{n+1}=^{\text{assumption of inductive step}}\\&= \frac {(3^{n+1})-1}{2}+3^{n+1}=\frac{3^{n+1}-1+2\cdot3^{n+1}}{2}=\frac{3\cdot3^{n+1}-1}{2}=\frac{3^{n+2}-1}{2}\\&=\frac{3^{(n+1)+1}-1}{2}\end{align*}$$ which proves that the expression holds also for $n+1$.

score 0 · Answer 2 · answered Apr 01 '14 at 08:28

You have the base step.

Induction hypothesis: Assume that $$\sum_{i=0}^k3^i=\frac{3^{k+1}-1}{2}$$

for some $k\ge0$

Inductive step: Then $$\begin{align}\sum_{i=0}^{k+1}&=3^{k+1}+\sum_{i=0}^k3^i\\&=3^{k+1}+\frac{3^{k+1}-1}{2}\\&=\frac{3^{k+2}-1}{2}\end{align}$$

Thus our formula is proved.

score 0 · Answer 3 · answered Apr 01 '14 at 08:30

0

Hint:$$\sum_{i=0}^{n+1}3^{i}=\sum_{i=0}^{n}3^{i}+3^{n+1}=\frac{3^{n+1}-1}{3-1}+3^{n+1}=\dots$$

answered Apr 01 '14 at 08:30

drhab

151,093

Prove by induction:$\sum_{i=0}^n 3^i = \frac {(3^{n+1})-1}{2}$

3 Answers3