Answer
$\displaystyle (1+ \frac{1}{x})^{6}=1+\frac{6}{x}+\frac{15}{x^{2}}+\frac{20}{x^{3}}+\frac{15}{x^{4}}+\frac{6}{x^{5}}+\frac{1}{x^{6}}$
Or written with negative exponents:
$\displaystyle (1+ \frac{1}{x})^{6}=1+6x^{-1}+15x^{-2}+20x^{-3}+15x^{-4}+6x^{-5}+x^{-6}$
Work Step by Step
We expand using the Binomial Theorem:
$\displaystyle (1+ \frac{1}{x})^{6}=\left(\begin{array}{l} 6\\ 0 \end{array}\right)1^{6}+\left(\begin{array}{l} 6\\ 1 \end{array}\right)1^{5}(\frac{1}{x})^1+\left(\begin{array}{l} 6\\ 2 \end{array}\right)1^{4}(\frac{1}{x})^{2}+\left(\begin{array}{l} 6\\ 3 \end{array}\right)1^{3}(\frac{1}{x})^{3}+\left(\begin{array}{l} 6\\ 4 \end{array}\right)1^{2}(\frac{1}{x})^{4}+\left(\begin{array}{l} 6\\ 5 \end{array}\right)1^{1}(\frac{1}{x})^{5}+\left(\begin{array}{l}
6\\ 6 \end{array}\right)(\frac{1}{x})^{6}=1+\frac{6}{x}+\frac{15}{x^{2}}+\frac{20}{x^{3}}+\frac{15}{x^{4}}+\frac{6}{x^{5}}+\frac{1}{x^{6}}$
Or written with negative exponents:
$=1+6x^{-1}+15x^{-2}+20x^{-3}+15x^{-4}+6x^{-5}+x^{-6}$
