Difference between revisions of "1959 AHSME Problems/Problem 39"
Clarkculus (talk | contribs) (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...") |
Clarkculus (talk | contribs) m (→Solution) |
||
Line 4: | Line 4: | ||
<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 12:31, 9 April 2024
Problem 37
Let S be the sum of the first nine terms of the sequence Then S equals:
Solution
We know . The only answer with this term is .
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 38 |
Followed by Problem 40 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.