fractions with powers

**Jskid** · November 28th 2011, 11:34 PM

How does $1-\frac{1}{3^{k-1}}+\frac{2}{3^k}=1- \frac{1}{3^k}$ ?
I tried putting it over a common denominator and got $1-\frac{3^k+6^{k-1}}{3^{k-1} \cdot 3^k}$

**princeps** · November 29th 2011, 12:27 AM

Originally Posted by Jskid

How does $1-\frac{1}{3^{k-1}}+\frac{2}{3^k}=1- \frac{1}{3^k}$ ?
I tried putting it over a common denominator and got $1-\frac{3^k+6^{k-1}}{3^{k-1} \cdot 3^k}$

$1-\frac{1}{3^{k-1}}+\frac{2}{3^k}=1-\frac{1}{3^{k-1}}+\frac{2}{3\cdot 3^{k-1}}=1-\left(\frac{1}{3^{k-1}}-\frac{2}{3\cdot 3^{k-1}}\right)=$

$=1-\left(\frac{3-2}{3\cdot 3^{k-1}}\right)=1-\frac{1}{3^k}$

**Soroban** · November 29th 2011, 04:17 AM

Hello, Jskid!

A slightly different approach . . .

$\text{How does }\,1-\frac{1}{3^{k-1}}+\frac{2}{3^k}\:=\:1- \frac{1}{3^k}\:?$

Consider the two fractions: . $- \frac{1}{3^{k-1}} + \frac{2}{3^k}$

Multiply the first fraction by $\tfrac{3}{3}\!:\;\;-\frac{1}{3^{k-1}}\cdot\frac{3}{3}\;+\;\frac{2}{3^k}$

Then we have: . $-\frac{3}{3^k} + \frac{2}{3^k} \;=\;\frac{-3 + 2}{3^k} \;=\;-\frac{1}{3^k}$

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