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Evaluate the sum (²¹C₁-¹⁰C₁)+(²¹C₂-¹⁰C₂)+(²¹C₃-¹⁰C₃)+(²¹C₄-¹⁰C₄)+⋯+(²¹C₁₀-¹⁰C₁₀).

1. 2²⁰-2¹⁰
2. 2²¹-2¹⁰
3. 2²¹-2¹¹
4. 2²⁰-2⁹

Correct answer: 2²⁰-2¹⁰

Solution

Sum C(21,k), k=1..10 = 2^20 - 1 (half of 2^21 minus the k=0 term). Sum C(10,k), k=1..10 = 2^10 - 1. Difference = (2^20 - 1) - (2^10 - 1) = 2^20 - 2^10.

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