# Difference between revisions of "2005 Alabama ARML TST Problems/Problem 7"

Mathgeek2006 (talk | contribs) m (→Solution) |
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We can compute those sums: | We can compute those sums: | ||

− | < | + | <cmath>\begin{eqnarray*} |

\sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)=x\\ | \sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)=x\\ | ||

=3\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\\ | =3\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\\ | ||

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z=4+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\cdots=6\\ | z=4+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\cdots=6\\ | ||

3+4+6=\boxed{13} | 3+4+6=\boxed{13} | ||

− | \end{eqnarray}</ | + | \end{eqnarray*}</cmath> |

==See Also== | ==See Also== |