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Toolkit + CCSS Solution
Understand
Restated: There is a sequence of numbers. For every positive whole number $n$, when you average the first $n$ terms, the average comes out to exactly $n$. Find the value of the 2008th term.
Givens: For each positive integer $n$, the mean (average) of the first $n$ terms equals $n$.; The average of the first 1 term is 1, of the first 2 terms is 2, of the first 3 terms is 3, and so on.
Unknowns: The value of the 2008th term of the sequence.
Understand
Restated: There is a sequence of numbers. For every positive whole number $n$, when you average the first $n$ terms, the average comes out to exactly $n$. Find the value of the 2008th term.
Givens: For each positive integer $n$, the mean (average) of the first $n$ terms equals $n$.; The average of the first 1 term is 1, of the first 2 terms is 2, of the first 3 terms is 3, and so on.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #5 Look for a Pattern
The average of a group equals its total divided by how many things are in it. So I give the running total a name, $S_n$, the sum of the first $n$ terms. The average rule turns into a clean formula for $S_n$. Then a single term is just the jump from one running total to the next, which splits the task into a small subproblem I can compute.
Execute — Answer: B
#4 Introduce a Variable 6.SP.B.5Step 1
Turn the average into a sum
Average = total ÷ count, so total = average × count.
Let $S_n$ be the sum of the first $n$ terms.
The mean of the first $n$ terms is $n$, and there are $n$ of them, so $S_n = n \times n = n^2$.
$$S_n = (\text{mean}) \times n = n \times n = n^2$$
💡 If a pile of $n$ numbers averages to $n$, the whole pile must add up to $n$ copies of $n$.
#7 Identify Subproblems 6.EE.B.6Step 2
One term is the jump between totals
The 2008th term is what gets added when the running total grows from the first 2007 terms to the first 2008 terms.
💡 The gap between two neighbouring squares is just the sum of the two numbers being squared.
[1]
#4 6.SP.B.5Average = total ÷ count, so total = average × count. Let $S_n$ be the sum of the
[2]
#7 6.EE.B.6The 2008th term is what gets added when the running total grows from the first 2
[3]
#5 6.EE.A.3Instead of computing two huge squares, note that $a^2 - b^2 = (a-b)(a+b)$. Here
Review
Reasonableness: Check the pattern on small cases. $S_1 = 1$, so term 1 is 1. $S_2 = 4$, so term 2 is $4-1=3$. $S_3 = 9$, so term 3 is $9-4=5$. The terms are $1, 3, 5, 7, \dots$, the odd numbers, and the $n$th odd number is $2n-1$. For $n=2008$ that gives $2(2008)-1 = 4015$, matching (B).
Alternative: You could skip the factoring and multiply directly: $2008^2 = 4{,}032{,}064$ and $2007^2 = 4{,}028{,}049$, and their difference is $4015$. The trap answers (D) and (E) are exactly those two squares, offered to catch anyone who computes a square but forgets to subtract.
CCSS standards used (min grade 6)
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures (Using the definition of mean (total ÷ count) to convert the average rule into the sum formula $S_n = n^2$.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming the running total $S_n$ and writing the 2008th term as $S_{2008} - S_{2007}$.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Rewriting $2008^2 - 2007^2$ as $(2008-2007)(2008+2007)$ to compute it quickly.)
⭐ If the average of the first $n$ terms is always $n$, the running total is $n^2$, so each term is the gap between two neighbouring squares.
⭐ If the average of the first $n$ terms is always $n$, the running total is $n^2$, so each term is the gap between two neighbouring squares.