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Binomial theorem (Milind Charakborty's question at Yahoo! Answers)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Milind Charakborty,

According to the binomial theorem: $$(a+b)^n=\displaystyle\sum_{i=0}^n{}\displaystyle\binom{n}{k}a^{n-k}b^k=\displaystyle\binom{n}{0}a^{n}+\displaystyle\binom{n}{1}a^{n-1}b^{}+\displaystyle\binom{n}{2}a^{n-2}b^{2}+\ldots\\+\displaystyle\binom{n}{n-3}a^{3}b^{n-3}+\displaystyle\binom{n}{n-2}a^{2}b^{n-1}+\displaystyle\binom{n}{n-1}a^{}b^{n-1}+\displaystyle\binom{n}{n}b^{n}$$ Using $\displaystyle\binom{n}{p}=\displaystyle\binom{n}{n-p}$: $$(a+b)^n=\ldots+\displaystyle\binom{n}{3}a^{3}b^{n-3}+\displaystyle\binom{n}{2}a^{2}b^{n-1}+\displaystyle\binom{n}{1}a^{}b^{n-1}+\displaystyle\binom{n}{0}b^{n}\\=\ldots +\frac{n(n-1)(n-2)}{3!}a^{3}b^{n-3}+\frac{n(n-1)}{2!}a^{2}b^{n-2}+na^{}b^{n-1}+b^{n}$$