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Difference between revisions of "1962 AHSME Problems/Problem 32"

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==Problem==
 
==Problem==
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If <math>x_{k+1} = x_k + \frac12 for k=1, 2, \dots, n-1</math> and <math>x_1=1</math>, find <math>x_1 + x_2 + \dots + x_n</math>.  
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If <math>x_{k+1} = x_k + \frac12</math> for <math>k=1, 2, \dots, n-1</math> and <math>x_1=1</math>, find <math>x_1 + x_2 + \dots + x_n</math>.  
  
 
<math> \textbf{(A)}\ \frac{n+1}{2}\qquad\textbf{(B)}\ \frac{n+3}{2}\qquad\textbf{(C)}\ \frac{n^2-1}{2}\qquad\textbf{(D)}\ \frac{n^2+n}{4}\qquad\textbf{(E)}\ \frac{n^2+3n}{4} </math>
 
<math> \textbf{(A)}\ \frac{n+1}{2}\qquad\textbf{(B)}\ \frac{n+3}{2}\qquad\textbf{(C)}\ \frac{n^2-1}{2}\qquad\textbf{(D)}\ \frac{n^2+n}{4}\qquad\textbf{(E)}\ \frac{n^2+3n}{4} </math>
  
 
==Solution==
 
==Solution==
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{{solution}}
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The sequence <math>x_1, x_2, \dots, x_n</math> is an arithmetic sequence since every term is <math>\frac12</math> more than the previous term. Letting <math>a=1</math> and <math>r=\frac12</math>, we can rewrite the sequence as <math>a, a+r, \dots, a+(n-1)r</math>.
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Recall that the sum of the first <math>n</math> terms of an arithmetic sequence is <math>na+\binom{n}2r</math>.
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Substituting our values for <math>a</math> and <math>r</math>, we get <math>n+\frac{\binom{n}2}{2}</math>. Simplifying gives
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<math>\boxed{\frac{n^2+3n}{4}\textbf{ (E)}}</math>

Latest revision as of 19:50, 31 May 2018

Problem

If $x_{k+1} = x_k + \frac12$ for $k=1, 2, \dots, n-1$ and $x_1=1$, find $x_1 + x_2 + \dots + x_n$.

$\textbf{(A)}\ \frac{n+1}{2}\qquad\textbf{(B)}\ \frac{n+3}{2}\qquad\textbf{(C)}\ \frac{n^2-1}{2}\qquad\textbf{(D)}\ \frac{n^2+n}{4}\qquad\textbf{(E)}\ \frac{n^2+3n}{4}$

Solution

The sequence $x_1, x_2, \dots, x_n$ is an arithmetic sequence since every term is $\frac12$ more than the previous term. Letting $a=1$ and $r=\frac12$, we can rewrite the sequence as $a, a+r, \dots, a+(n-1)r$. Recall that the sum of the first $n$ terms of an arithmetic sequence is $na+\binom{n}2r$. Substituting our values for $a$ and $r$, we get $n+\frac{\binom{n}2}{2}$. Simplifying gives $\boxed{\frac{n^2+3n}{4}\textbf{ (E)}}$

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