# How do you find a_10 given a_n=2^n/(n!)?

$\frac{4}{14 , 175}$

#### Explanation:

We have the general expression a_n=2^n/(n!) and are looking for ${a}_{10}$. That gives us:

a_10=2^10/(10!)

And now let's evaluate it! (I'll express the numbers 8 and 4 in terms of ${2}^{x}$ to make cancellations and simplification more obvious):

$= \frac{{2}^{3} \times {2}^{2} \times 2 \times 2 \times 2 \times {2}^{2}}{10 \times 9 \times {2}^{3} \times 7 \times 6 \times 5 \times {2}^{2} \times 3 \times 2}$

$= \frac{\cancel{\textcolor{red}{{2}^{3}}} \times \cancel{\textcolor{b r o w n}{{2}^{2}}} \times \cancel{\textcolor{\mathmr{and} a n \ge}{2}} \times \cancel{\textcolor{g r e e n}{2}} \times \cancel{\textcolor{b l u e}{2}} \times {2}^{2}}{{\cancel{\textcolor{g r e e n}{10}}}^{5} \times 9 \times \cancel{\textcolor{red}{{2}^{3}}} \times 7 \times {\cancel{\textcolor{b l u e}{6}}}^{3} \times 5 \times \cancel{\textcolor{b r o w n}{{2}^{2}}} \times 3 \times \cancel{\textcolor{\mathmr{and} a n \ge}{2}}}$

$= \frac{4}{5 \times 9 \times 7 \times 3 \times 5 \times 3} = \frac{4}{14 , 175}$