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(a+b)n=∑k=0n(nk)an−kbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k(a+b)n=∑k=0n(kn)an−kbk
(n,k=)n!k!(n−k)!\binom{n, k} = \frac{n!}{k!(n-k)!}(=n,k)k!(n−k)!n!
(1+x)n=∑k=0∞(nk)xk(1 + x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k(1+x)n=∑k=0∞(kn)xk
It converges for ∣x∣<1|x| < 1∣x∣<1.
The second term is nxnxnx for (1+x)n(1 + x)^n(1+x)n, n≥1n \geq 1n≥1.
Use multinomial expansion.
∑k=0n∑l=0k(nk)(kl)xlyk−l\sum_{k=0}^{n} \sum_{l=0}^{k} \binom{n}{k} \binom{k}{l} x^l y^{k-l}∑k=0n∑l=0k(kn)(lk)xlyk−l
To approximate functions using polynomial expressions.
1+12x−18x2+116x3+...1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + ...1+21x−81x2+161x3+...
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