1959 AMC 12 Problem 39

Question

Let

S

be the sum of the first nine terms of the sequence

\begin{gathered} x+a,\quad x^2+2a,\\ x^3+3a,\quad\ldots. \end{gathered}

Then

S

equals:

\dfrac{50a+x+x^8}{x+1}

50a-\dfrac{x+x^{10}}{x-1}

\dfrac{x^9-1}{x+1}+45a

\dfrac{x^{10}-x}{x-1}+45a

\dfrac{x^{11}-x}{x-1}+45a

Separate the powers of

x

from the multiples of

a

Use the geometric-series sum for

x+x^2+\cdots+x^9

Adding the first nine terms gives

\begin{aligned} S={}&(x+x^2+\cdots+x^9)\\ &{}+(1+2+\cdots+9)a. \end{aligned}

Therefore

S=\frac{x^{10}-x}{x-1}+45a.

The expression is understood by continuity at

x=1,

where both forms equal

9+45a.

Thus, the correct answer is D.

Problem 39 in other years