1962 AHSME Problems/Problem 32: Difference between revisions
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==Problem== | ==Problem== | ||
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>. | 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== | ||
{{ | 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>. | ||
Recall that the sum of the first <math>n</math> terms of an arithmetic sequence is <math>na+\binom{n}2b</math>. | |||
Substituting our values for <math>a</math> and <math>r</math>, we get <math>n+\frac{\binom{n}2}2}</math>. Simplifying gives | |||
<math>\boxed{\frac{n^2+3n}{4}\textbf{ (E)}}</math> | |||
Revision as of 20:24, 16 April 2014
Problem
If 
for 
and 
, find 
.
Solution
The sequence 
is an arithmetic sequence since every term is 
more than the previous term. Letting 
and 
, we can rewrite the sequence as 
.
Recall that the sum of the first 
terms of an arithmetic sequence is 
.
Substituting our values for 
and 
, we get $n+\frac{\binom{n}2}2}$ (Error compiling LaTeX. Unknown error_msg). Simplifying gives
