Logo for LIVE by Po-Shen Loh

LIVE

English

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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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
Difficulty rating: 1440
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.

← Problem 31#31
Full Exam
Problem 33#33 →

Problem 32 in Other Years

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12 · 1960 AMC 12 · 1961 AMC 12 · 1963 AMC 12 · 1964 AMC 12 · 1965 AMC 12 · 1966 AMC 12 · 1967 AMC 12 · 1968 AMC 12 · 1969 AMC 12 · 1970 AMC 12 · 1971 AMC 12 · 1972 AMC 12 · 1973 AMC 12