# Exponents in the denominator?

I'm having trouble understanding exponents in the denominator.

For example: I have the expression: $\displaystyle 1 - \frac{1}{3^n} + \frac{2}{3^{n+1}}$. I know that this simplifies to $\displaystyle 1 - \frac{1}{3^{n+1}}$, but how/why? Can someone please list the steps?

My understanding is that the exponent $(n+1)$ in the expression $x^{n+1}$ means that $x^{n+1} = x x^n$, but how does this fit with the above problem?

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You should get a common denominator for the last two terms by multiplying the second term by $3/3$. You have \begin{align*} 1-\frac{1}{3^n}+\frac{2}{3^{n+1}} &= 1-\frac{3}{3^{n+1}}+\frac{2}{3^{n+1}}\\ &= 1-\frac{1}{3^{n+1}} \end{align*}

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My understanding is that the exponent $n+1$ means that $x^{n+1}=x\cdot x^n$

Your understanding is correct, and you can apply it to the problem at hand by remembering that when an exponent, say $n$, of an expression is in the denominator of a fraction, it can be written as an expression raised to the $-n$ power. Then you can manipulate exponents by the respective rules to which you are accustomed. $$\frac{1}{x^{n+1}} = x^{-(n+1)} = x^{-n-1} = x^{-1}x^{-n} = \frac {1}{x\cdot x^n} < \frac{1}{x^n} = x^{-n}$$(Note: the inequality holds, as stated, provided $n > 1$.)

That is, so we have that the "least common denominator" is $x\cdot x^n$, with $$\frac{1}{x^n} = x^{-n} = x^1\cdot x^{-n - 1} = \frac{x}{x\cdot x^n} = \frac {x}{x^{n+1}}$$

In your problem, then $$1-\frac{1}{3^n}+\frac{2}{3^{n+1}} = 1 - 3^{-n} + 2\cdot 3^{-(n+1)} = 1 - 3\cdot 3^{-n - 1} + 2\cdot 3^{-n - 1}$$ $$= 1 + (-3 + 2)\cdot 3^{-n-1} = 1 - 3^{-(n+1)} = 1 - \frac{1}{3^{n+1}}$$

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