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Difference between revisions of "1959 AHSME Problems/Problem 39"

(Created page with "== Problem 37== Let S be the sum of the first nine terms of the sequence <math>x+a, x^2+2a, x^3+3a, \cdots.</math> Then S equals: <math>\textbf{(A)}\ \frac{50a+x+x^8}{x+1} \qq...")
 
m (→‎Solution)
 
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<math>\textbf{(A)}\ \frac{50a+x+x^8}{x+1} \qquad\textbf{(B)}\ 50a-\frac{x+x^{10}}{x-1}\qquad\textbf{(C)}\ \frac{x^9-1}{x+1}+45a\qquad\textbf{(D)}\ \frac{x^{10}-x}{x-1}+45a\qquad\textbf{(E)}\ \frac{x^{11}-x}{x-1}+45a  </math>   
 
<math>\textbf{(A)}\ \frac{50a+x+x^8}{x+1} \qquad\textbf{(B)}\ 50a-\frac{x+x^{10}}{x-1}\qquad\textbf{(C)}\ \frac{x^9-1}{x+1}+45a\qquad\textbf{(D)}\ \frac{x^{10}-x}{x-1}+45a\qquad\textbf{(E)}\ \frac{x^{11}-x}{x-1}+45a  </math>   
 
== Solution ==
 
== Solution ==
−
We know <math>x+x^2\dots+x^9=x(1+x\dots+x^8)=x\frac{x^9-1}{x-1}=\frac{x^10-x}{x-1}</math>. The only answer with this term is <math>\boxed{\textbf{D}}</math>
+
We know <math>x+x^2\dots+x^9=x(1+x\dots+x^8)=x\frac{x^9-1}{x-1}=\frac{x^{10}-x}{x-1}</math>. The only answer with this term is <math>\boxed{\textbf{D}}</math>.
 +
 
 
== See also ==
 
== See also ==
 
{{AHSME 50p box|year=1959|num-b=38|num-a=40}}
 
{{AHSME 50p box|year=1959|num-b=38|num-a=40}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:31, 9 April 2024

Problem 37

Let S be the sum of the first nine terms of the sequence $x+a, x^2+2a, x^3+3a, \cdots.$ Then S equals: $\textbf{(A)}\ \frac{50a+x+x^8}{x+1} \qquad\textbf{(B)}\ 50a-\frac{x+x^{10}}{x-1}\qquad\textbf{(C)}\ \frac{x^9-1}{x+1}+45a\qquad\textbf{(D)}\ \frac{x^{10}-x}{x-1}+45a\qquad\textbf{(E)}\ \frac{x^{11}-x}{x-1}+45a$

Solution

We know $x+x^2\dots+x^9=x(1+x\dots+x^8)=x\frac{x^9-1}{x-1}=\frac{x^{10}-x}{x-1}$. The only answer with this term is $\boxed{\textbf{D}}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 38 		Followed by
Problem 40
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All AHSME Problems and Solutions

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