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.

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- Jan 20th 2009, 10:18 PMazukiMathematic induction Question 2
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. - Jan 20th 2009, 10:36 PMJhevon
We proceed by mathematical induction.

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

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

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

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

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

Then

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

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

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