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Finding Coefficients Quickly in the TMUA
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303 回視聴19高評価2:33JPiMaths元のリリース: 2026-05-16

To find the coefficient of x^5 in the expansion of (1 + x + x^2 + x^3 + x^4 + x^5)(1 + x)^5, recognize that the first polynomial is a geometric series summing to (1 - x^6)/(1 - x), and the second is the binomial expansion of (1 + x)^5. The coefficient of x^5 in the product equals the sum of binomial coefficients from (1 + x)^5, which is 2^5 = 32. This demonstrates that the sum of binomial coefficients (n choose 0) + (n choose 1) + ... + (n choose n) equals 2^n.

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