Problem 9 · AMC 8 Stretch
Stretch
Algebra & Patterns
Counting & Probability
reduce-and-expandpattern-recognition
When you multiply out \((x+y)^4\), you get \(x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4\). The numbers out front are \(1, 4, 6, 4, 1\), and they add up to \(16\). Without multiplying anything out, find the sum of the front numbers for \((x+y)^8\).
Show answer
Answer: \(2^8 = 256\)
Show hints
Hint 1 of 3
Here is the trick: the 'sum of the front numbers' is just what you get if you plug in \(x=1\) and \(y=1\), because then every \(x\) and \(y\) turns into \(1\) and each term becomes only its number.
Still stuck? Show hint 2 →
Hint 2 of 3
Check it on the small example: \((1+1)^4 = 2^4 = 16\), which matches \(1+4+6+4+1\). Cool!
Still stuck? Show hint 3 →
Hint 3 of 3
So for \((x+y)^8\), just compute \((1+1)^8 = 2^8\).
Show solution
Approach: Reduce and expand — set the variables to 1
- To add up the front numbers (coefficients), set \(x=1\) and \(y=1\). Then every term becomes just its front number, so the whole expression equals the sum of those numbers.
- Check on the warm-up: \((1+1)^4 = 2^4 = 16\), and indeed \(1+4+6+4+1 = 16\).
- Do the same for the eighth power: \((1+1)^8 = 2^8 = 256\). So the front numbers of \((x+y)^8\) add up to \(256\) — no multiplying out required.
Mark:
· log in to save