Essential Formulae Binomial expansion \({{\left( a+b \right)}^{n}}={{a}^{n}}+\left( \begin{matrix} n

Question

Essential Formulae Binomial expansion ${{\left( a+b \right)}^{n}}={{a}^{n}}+\left( \begin{matrix} n \ 1 \ \end{matrix} \right){{a}^{n-1}}b+\left( \begin{matrix} n \ 2 \ \end{matrix} \right){{a}^{n-2}}{{b}^{2}}+...+\left( \begin{matrix} n \ r \ \end{matrix} \right){{a}^{n-r}}{{b}^{r}}+...+{{b}^{n}}$, where $n$ is a positive integer and $\left( \begin{matrix} n \ r \ \end{matrix} \right)=\frac{n!}{r!\left( n-r \right)!}=\frac{n\left( n-1 \right)...\left( n-r+1 \right)}{r!}$. Formula for $\left( \begin{matrix} n \ r \ \end{matrix} \right)$or ${}^{n}{{C}_{r}}$ $\left( \begin{matrix} n \ r \ \end{matrix} \right)=\frac{n!}{r!\left( n-r \right)!}=\frac{n\left( n-1 \right)...\left( n-r+1 \right)}{r!}$ E.g. $\left( \begin{matrix} 10 \ 2 \ \end{matrix} \right)=\frac{10!}{8!2!}=\frac{10\times 9}{2!}$, $\left( \begin{matrix} 7 \ 3 \ \end{matrix} \right)=\frac{7!}{4!3!}=\frac{7\times 6\times 5\times 4}{3!}$ Special Identities to Remember • $0!=1$ • $\left( \begin{matrix} n \ 0 \ \end{matrix} \right)=1$, $\left( \begin{matrix} n \ 1 \ \end{matrix} \right)=n$, $\left( \begin{matrix} n \ 2 \ \end{matrix} \right)=\frac{n(n-1)}{2!}$, • $\left( \begin{matrix} n \ 3 \ \end{matrix} \right)=\frac{n(n-1)(n-2)}{3!}$, $\left( \begin{matrix} n \ n \ \end{matrix} \right)=1$ • $\left( \begin{matrix} n \ r \ \end{matrix} \right)=\left( \begin{matrix} n \ n-r \ \end{matrix} \right)$ • $\left( \begin{matrix} n \ 0 \ \end{matrix} \right)+\left( \begin{matrix} n \ 1 \ \end{matrix} \right)+\ldots +\left( \begin{matrix} n \ n-1 \ \end{matrix} \right)+\left( \begin{matrix} n \ n \ \end{matrix} \right)={{2}^{n}}$

Accepted Answer

$(a+b)^n=\sum_{r=0}^{n}\binom nr a^{n-r}b^r$, where $\binom nr=\frac{n!}{r!(n-r)!}$ and $T_{r+1}=\binom nr a^{n-r}b^r$

Binomial Theorem - Binomial Coefficients and Pascal's Triangle

Binomial Theorem - Binomial Coefficients and Pascal's Triangle

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