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2008 AMC 10B Problem 13

2008 AMC 10B Problem 13AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 2 of 5 · about 2 min · position P11-15

For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the 2008th term of the sequence?

A)
${{{2008}}}$
B)
${{{4015}}}$
C)
${{{4016}}}$
D)
${{{4,030,056}}}$
E)
${{{4,032,064}}}$

Answer

Key insight

Mean n over n terms means the partial sums are n^2, so the n-th term is n^2 - (n-1)^2 = 2n - 1.

Solution

Let $S_n$ be the sum of the first $n$ terms. A mean of $n$ over $n$ terms means
$S_n = n \cdot n = n^2.$
The $2008$ th term is the amount the sum grows when that term is added:
$a_{2008} = S_{2008} - S_{2007} = 2008^2 - 2007^2 = (2008 - 2007)(2008 + 2007) = 4015.$

The answer is $\boxed{\textbf{(B)}\ 4015}$ .

Why this works

Information about averages of the first $n$ terms is really information about partial sums, and any single term is a difference of two consecutive partial sums. Here $S_n = n^2$ produces the odd numbers $a_n = 2n - 1$ , a pattern worth remembering: the sum of the first $n$ odd numbers is $n^2$ .

Alternative approach

Compute the first few terms: $a_1 = 1$ , $a_2 = 4 - 1 = 3$ , $a_3 = 9 - 4 = 5$ . The terms are the odd numbers, so $a_{2008} = 2 \cdot 2008 - 1 = 4015$ .

The trap

Answering 2008 or 4016 by confusing the term with the mean or the partial-sum difference, instead of 2(2008) - 1.

Common mistakes

Answering 2008 or 4016 by confusing the term with the mean or the partial-sum difference, instead of 2(2008) - 1.
Reporting $S_{2008} = 2008^2 = 4{,}032{,}064$ , the sum rather than the term.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta