Difference between revisions of "2005 Alabama ARML TST Problems/Problem 7"
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==Solution== | ==Solution== | ||
| + | <math>\sum_{n=1}^{\infty} \frac{n^2+2n+3}{2^n}=\sum_{n=1}^{\infty} \left(\frac{n^2}{2^n}\right)+\sum_{n=1}^{\infty} \left(\frac{2n}{2^n}\right)+\sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)</math> | ||
| − | {{ | + | We can compute those sums: |
| + | |||
| + | <cmath>\begin{eqnarray*} | ||
| + | \sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)=x\\ | ||
| + | =3\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\\ | ||
| + | 2x=3\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\\ | ||
| + | x=3(1)=3\\ | ||
| + | \sum_{n=1}^{\infty} \left(\frac{2n}{2^n}\right)=y\\ | ||
| + | =\frac{2}{2}+\frac{4}{4}+\frac{6}{8}+\frac{8}{16}+\cdots\\ | ||
| + | 2y=2+\frac{4}{2}+\frac{6}{4}+\frac{8}{8}+\cdots\\ | ||
| + | y=2+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\cdots=4\\ | ||
| + | \sum_{n=1}^{\infty} \left(\frac{n^2}{2^n}\right)=z\\ | ||
| + | =\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\cdots\\ | ||
| + | 2z=1+\frac{4}{2}+\frac{9}{4}+\frac{16}{8}+\cdots\\ | ||
| + | z=1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\frac{9}{16}+\cdots\\ | ||
| + | 2z=2+3+\frac{5}{2}+\frac{7}{4}+\frac{9}{8}+\cdots\\ | ||
| + | z=4+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\cdots=6\\ | ||
| + | 3+4+6=\boxed{13} | ||
| + | \end{eqnarray*}</cmath> | ||
==See Also== | ==See Also== | ||
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