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Question Number 136783 by Abdoulaye last updated on 26/Mar/21

$$a,b\in R$$$${(a+b)}^n=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)a^k{b}^{n-k}$$$$demontration!$$

Commented by mr W last updated on 26/Mar/21

$${(a+b)}^n=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)a^{\mathrm{n}-\mathrm{k}}{b}^k=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)a^{\mathrm{k}}{b}^{n-k}$$

Commented by abdurehime last updated on 26/Mar/21

$$a,b\in R$$$${(a+b)}^n=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)a^{\mathrm{n}-\mathrm{k}}{b}^k$$$$demontration!\mathrm{this}\mathrm{the}\mathrm{correct}\mathrm{one}$$$$\mathrm{also}\mathrm{i}\mathrm{want}\mathrm{to}\mathrm{see}\mathrm{the}\mathrm{proof}$$

Answered by mr W last updated on 26/Mar/21

$$f\left(x\right)={(1+x)}^n$$$$f^{\left(1\right)}\left(x\right)=n{(1+x)}^{n-1}$$$$f^{\left(2\right)}\left(x\right)=n(n-1){(1+x)}^{n-2}$$$$f^{\left(k\right)}\left(x\right)=n(n-1)...(n-k+1){(1+x)}^{n-k}$$$$f^{\left(n\right)}\left(x\right)=n(n-1)...2\times 1$$$$f^{(n+...)}\left(x\right)=0$$$$f\left(x\right)=f\left(0\right)+f^{\left(1\right)}\left(0\right)x+f^{\left(2\right)}\left(0\right)\frac{x^2}{2!}+...+f^{\left(k\right)}\left(0\right)\frac{x^k}{k!}+...+f^{\left(n\right)}\left(o\right)\frac{x^n}{n!}+f^{(n+1)}\left(0\right)\frac{x^{n+1}}{(n+1)!}+...$$$$f\left(x\right)=1+nx+\frac{n(n-1)}{2!}{x}^2+...+\frac{n(n-1)...(n-k+1)}{k!}{x}^k+...+\frac{n!}{n!}{x}^n+0\times \frac{x^{n+1}}{(n+1)!}+...$$$$f\left(x\right)=1+nx+\frac{n(n-1)}{2!}{x}^2+...+\frac{n(n-1)...(n-k+1)}{k!}{x}^k+...+\frac{n!}{n!}{x}^n$$$$f\left(x\right)=\left(\begin{array}{c}n\\ 0\end{array}\right)x^0+\left(\begin{array}{c}n\\ 1\end{array}\right)x^1+\left(\begin{array}{c}n\\ 2\end{array}\right)x^2+...+\left(\begin{array}{c}n\\ k\end{array}\right)x^k+...+\left(\begin{array}{c}n\\ n\end{array}\right)x^n$$$$f\left(x\right)={(1+x)}^n=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)x^k$$$$a^n{(1+x)}^n=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)a^n{x}^k$$$$letx=\frac{b}{a}$$$$a^n{(1+\frac{b}{a})}^n=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)a^n{\left(\frac{b}{a}\right)}^k$$$$\Rightarrow {(a+b)}^n=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)a^{n-k}{b}^k$$$$or$$$$b^n{(1+\frac{a}{b})}^n=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)b^n{\left(\frac{a}{b}\right)}^k$$$$\Rightarrow {(a+b)}^n=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)a^k{b}^{n-k}$$

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