二項式定理(Binomial Theorem)

\[\binom{r}{n}=\frac{n!}{r!\left(n-r\right)!}\]

證明

\[Let\ P(n)\ be\ a\ statement\ \left(a+b\right)^n=\sum_{r=0}^n\binom{r}{k}a^rb^{n-r}\ \ \ \ \ where\ \ n\ is\ a\ positive\ integer\]

\[when\ n=1\]

\[L.S.=\left(a+b\right)\]

\[R.S.=\sum_{r=0}^1\binom{1}{r}a^rb^{1-r}=\binom{1}{0}a^0b^{1-0}+\binom{1}{1}a^1b^{1-1}=a+b=L.S.\]

\[P\left(1\right)\ holds\]

\[Assume\ P\left(k\right)\ holds\]

\[i.e.\ \ \ \ \left(a+b\right)^k=\sum_{r=0}^k\binom{k}{r}a^rb^{k-r}\]

\[When\ n=k+1\]

\[\left(a+b\right)^{k+1}=\left(a+b\right)^k\left(a+b\right)\]

\[=\left(\sum_{r=0}^k\binom{k}{r}a^rb^{k-r}\right)\left(a+b\right)\]

\[=\sum_{r=0}^k\binom{k}{r}a^{r+1}b^{k-r}+\sum_{r=0}^k\binom{k}{r}a^rb^{k-r+1}\]

\[=a^{k+1}+\sum_{r=0}^{k-1}\binom{k}{r}a^{r+1}b^{k-r}+\sum_{r=1}^{k-1}\binom{k}{r}a^rb^{k-r+1}+b^{k+1}\]

\[=a^{k+1}+\sum_{r=0}^{k-1}\binom{k}{r}a^{r+1}b^{k-r}+\sum_{r=0}^{k-1}\binom{k}{r+1}a^{r+1}b^{k-r}+b^{k+1}\]

\[=a^{k+1}+\sum_{r=0}^{k-1}\left(\binom{k}{r}a^{r+1}b^{k-r}+\binom{k}{r+1}a^{r+1}b^{k-r}\right)+b^{k+1}\]

\[=a^{k+1}+\sum_{r=0}^{k-1}\left(\left(\binom{k}{r}+\binom{k}{r+1}\right)a^{r+1}b^{k-r}\right)+b^{k+1}\]

\[=a^{k+1}+\sum_{r=0}^{k-1}\binom{k+1}{r+1}a^{r+1}b^{k-r}+b^{k+1}\]

\[=a^{k+1}+\sum_{r=1}^{k-1}\binom{k+1}{r}a^rb^{k-r+1}+b^{k+1}\]

\[=\binom{k+1}{0}a^0b^{k-0+1}+\sum_{r=1}^k\binom{k+1}{r}a^rb^{k-r+1}=\sum_{r=0}^k\binom{k+1}{r}a^rb^{\left(k+1\right)-r}\]

\[P\left(k\right)\Rightarrow P\left(k+1\right)\]

\[By\ mathematical\ induction,\ P\left(n\right)\ is\ true\ \]

\[\left(a+b\right)^n=\sum_{r=0}^n\binom{n}{r}a^rb^{n-r}\ \ \ \ \ \ \ where\ \ n\ is\ a\ positive\ integer\]