Mathematic induction Question 2

**azuki** · January 20th 2009, 10:18 PM

Prove this by MI.
∑_(r=1)^3▒1/3^r = 1/2 - 1/(2(3^n))
How do I display this properly? I write from MS words and paste here. I try wrapping with Math tag but it wont work.

**Jhevon** · January 20th 2009, 10:36 PM

Originally Posted by azuki

Prove this by MI.
∑_(r=1)^3▒1/3^r = 1/2 - 1/(2(3^n))

We proceed by mathematical induction.

Let $P(n):~\sum_{r = 1}^n \frac 1{3^r} = \frac 12 - \frac 1{2 \cdot 3^n}$ for all $n \in \mathbb{N},~n \ge 1$

$P(1)$ is true, since $\sum_{r = 1}^1 \frac 1{3^r} = \frac 13 = \frac 12 - \frac 1{2 \cdot 3}$

Assume $P(n)$ is true, we now show that this implies $P(n + 1)$ is true.

Since $P(n)$ is true, we have

$\sum_{r = 1}^n \frac 1{3^r} = \frac 12 - \frac 1{2 \cdot 3^r}$

Then

$P(n + 1):~ \sum_{r = 1}^{n + 1} \frac 1{3^r} = \sum_{r = 1}^n \frac 1{3^r} + \frac 1{3^{n + 1}}$

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i leave it to you to finish up. my last line was a crucial hint

How do I display this properly? I write from MS words and paste here. I try wrapping with Math tag but it wont work.

see here

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