2008 AMC 10B Problem 13

Attempt Problem 13 of the 2008 AMC 10B 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 2008 AMC 10B solutions, or check the answer key.

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

13.

For each positive integer

n,

the mean of the first

n

terms of a sequence is

n.

What is the

2008

th term of the sequence?

2008

4015

4016

4{,}030{,}056

4{,}032{,}064

The sum of the first

n

terms is

n\cdot n=n^2

The

n

th term is

n^2-(n-1)^2

Since the mean of the first

n

terms is

n,

their sum is

n^2.

The

n

th term is

n^2-(n-1)^2=2n-1,

so the

2008

th term is

2\cdot 2008-1=4015.

Thus, the correct answer is B.

Problem 13 in other years