Problem 27 · 2010 Math Kangaroo
Stretch
Algebra & Patterns
difference-of-squaresgrouping
The expression \(\dfrac{(2+3)(2^2+3^2)\cdots(2^{1024}+3^{1024})(2^{2048}+3^{2048})+2^{4096}}{3^{2048}}\) is equal to
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Answer: C — \(3^{2048}\)
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Hint 1 of 2
Multiply the whole product by the missing factor (3 − 2) = 1; it telescopes.
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Hint 2 of 2
Each step uses (a−b)(a+b) = a² − b².
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Approach: telescoping difference of squares
- Since 3 − 2 = 1, the product equals (3−2)(3+2)(3²+2²)...(3^2048+2^2048) = 3^4096 − 2^4096.
- Adding 2^4096 gives 3^4096.
- Dividing by 3^2048 leaves 3^2048.
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