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Binomial Theorem

Use Pascal’s triangle and the binomial theorem to determine binomial coefficients. Expand $(a+b)^5$ .

Pascal’s Triangle

$(a+b)^0$ 1

$(a+b)^1$ 1 1

$(a+b)^2$ 1 2 1

$(a+b)^3$ 1 3 3 1

$(a+b)^4$ 1 4 6 4 1

$(a+b)^5$ 1 5 10 10 5 1

Binomial Theorem: If n is a natural number, then…

$(a+b)^{n}=nC_{0}a^{n}b^{0}+nC_{1}a^{n-1}b^{1}+nC_{2}a^{n-1}b^{2}+...+nC_{n}a^{0}b^{n}=\sum_{k=0}^{h} \frac{n!}{k!(n-k)!}a^{n-k}b^{k}$

$(a+b)^{5}=1a^{5}b^{0}+5a^{4}b^{1}+10a^{3}b^{2}+10a^{2}b^{3}+5a^{1}b^{4}+1a^{0}b^{5}$

Check your work:

$n=5$

$k=0:\ \frac{5!}{0!(5-0)!}a^{(5-0)}b^{0}=1a^{5}b^{0}$

$k=1:\ \frac{5!}{1!(5-1)!}a^{(5-1)}b^{1}=5a^{4}b^{1}$

$k=2:\ \frac{5!}{2!(5-2)!}a^{(5-2)}b^{2}=10a^{3}b^{2}$

$k=3:\ \frac{5!}{3!(5-3)!}a^{(5-3)}b^{3}=10a^{2}b^{3}$

$k=1:\ \frac{5!}{4!(5-4)!}a^{(5-4)}b^{4}=5a^{1}b^{4}$

$k=5:\ \frac{5!}{5!(5-5)!}a^{(5-5)}b^{5}=1a^{0}b^{5}$

$=1a^{5}b^{0}+5a^{4}b^{1}+10a^{3}b^{2}+10a^{2}b^{3}+5a^{1}b^{4}+1a^{0}b^{5}$