約 19,200,000 件の結果
- (1 + x)n = ∑k=0n (n k)xk (1 + x) n = ∑ k = 0 n (n k) x k for any positive integer n n, which is just the Taylor series for (1 + x)n (1 + x) n. This formula can be extended to all real powers α α: (1 + x)α = ∑k=0∞ (α k)xk (1 + x) α = ∑ k = 0 ∞ (α k) x kmath.hmc.edu/calculus/hmc-mathematics-calculus-online-tutorials/precalculus/b…
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Binomial expansion of $(1-x)^n$ - Mathematics Stack Exchange
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