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