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:

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

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

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

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

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

Answer: D

Concepts:geometric sequence arithmetic sequence summation

Difficulty rating: 1280

Small Hint:

Separate the powers of $x$ from the multiples of $a$

Big Hint:

Use the geometric-series sum for $x+x^2+\cdots+x^9$

Solution:

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.

1959 AMC 12 Problem 39

Answer: D

Difficulty rating: 1280

Small Hint:

Big Hint:

Solution:

Problem 39 in Other Years