1959 AHSME Problems/Problem 39

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 38
Followed by
Problem 40
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