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

(New page: ==Problem== Find the sum of the infinite series: <center><math>3+\frac{11}4+\frac 94 + \cdots + \frac{n^2+2n+3}{2^n}+\cdots</math>.</center> ==Solution== {{solution}} ==See Also== * [...)
 
(→‎Solution)
 
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==Solution==
 
==Solution==
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<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>
  
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{{solution}}
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We can compute those sums:
 +
 
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<cmath>\begin{eqnarray*}
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\sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)=x\\
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=3\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\\
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2x=3\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\\
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x=3(1)=3\\
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\sum_{n=1}^{\infty} \left(\frac{2n}{2^n}\right)=y\\
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=\frac{2}{2}+\frac{4}{4}+\frac{6}{8}+\frac{8}{16}+\cdots\\
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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\\
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z=4+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\cdots=6\\
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3+4+6=\boxed{13}
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\end{eqnarray*}</cmath>
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If you didn't understand the third case / summation, all they did was subtract the summation for <math>z</math> from <math>2z</math> to get the new <math>z</math> and then repeat this.
  
 
==See Also==
 
==See Also==
  
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* [[2005 Alabama ARML TST Problems]]
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{{ARML box|year=2005|state=Alabama|num-b=6|num-a=8}}
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* [[2005 Alabama ARML TST Problems/Problem 6|Previous Problem]]
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* [[2005 Alabama ARML TST Problems/Problem 8|Next Problem]]
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 20:13, 18 May 2021

Problem

Find the sum of the infinite series:

$3+\frac{11}4+\frac 94 + \cdots + \frac{n^2+2n+3}{2^n}+\cdots$.

Solution

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

We can compute those sums:

\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*}

If you didn't understand the third case / summation, all they did was subtract the summation for $z$ from $2z$ to get the new $z$ and then repeat this.

See Also

2005 Alabama ARML TST (Problems)
Preceded by:
Problem 6 		Followed by:
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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