1962 AMC 12 Problem 32

Attempt Problem 32 of the 1962 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1962 AMC 12 solutions, or check the answer key.

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32.

If xk+1=xk+12x_{k+1}=x_k+\dfrac12 for k=1,k=1, 2,2, …,\ldots, n−1n-1 and x1=1,x_1=1, find x1+x2+⋯+xn.x_1+x_2+\cdots+x_n.
n+12\dfrac{n+1}{2}
n+32\dfrac{n+3}{2}
n2−12\dfrac{n^2-1}{2}
n2+n4\dfrac{n^2+n}{4}
n2+3n4\dfrac{n^2+3n}{4}
Answer: E
Concepts:arithmetic sequencesummation
Small hint:
The recurrence defines an arithmetic sequence with common difference 12\frac{1}{2}
Big hint:
Find xnx_n, then use n(x1+xn)2\frac{n(x_1+x_n)}{2}
Solution:
We have xn=1+n−12=n+12.x_n=1+\frac{n-1}{2}=\frac{n+1}{2}. Therefore x1+⋯+xn=n2(1+n+12)=n2+3n4. \begin{aligned} x_1+\cdots+x_n &=\frac n2\left(1+\frac{n+1}{2}\right)\\ &=\frac{n^2+3n}{4}. \end{aligned}
Thus, the correct answer is E.
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