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Question Number 183459 by Matica last updated on 26/Dec/22

$$Itisgivenf\left(x\right)={(1+x)}^n,n\in \mathbb{N}.Find$$$$f\left(0\right)+f^{{}^{\prime }}\left(0\right)+\frac{f^{{}^{″}}\left(0\right)}{2!}+\frac{f^{‴}\left(0\right)}{3!}+...+\frac{f^{\left(n\right)}\left(0\right)}{n!}.$$

Answered by mahdipoor last updated on 26/Dec/22

$$=1+n+\frac{n\times (n-1)}{2}+\frac{n\times (n-1)\times (n-2)}{3!}+...$$$$\frac{n!}{n!}=\left({}_0^n\right)+\left({}_1^n\right)+...\left({}_n^n\right)=2^n$$

Commented by Matica last updated on 26/Dec/22

$$pleasemoredetail$$

Commented by mahdipoor last updated on 26/Dec/22

$$f\left(0\right)={(1+0)}^n=1=\left({}_0^n\right)$$$$f^{{}^{\prime }}\left(0\right)=n{(1+0)}^{n-1}=n=\left({}_1^n\right)$$$$\frac{1}{2!}{f}^{{}^{″}}\left(0\right)=\frac{n(n-1)}{2!}{(1+0)}^{n-2}=\left({}_2^n\right)$$$$....$$$$\frac{1}{n!}{f}^n\left(0\right)=\frac{n!}{n!}{(1+0)}^{n-n}=\left({}_n^n\right)$$$$\Rightarrow A=\left({}_0^n\right)+\left({}_1^n\right)+...\left({}_n^n\right)$$$$B={(a+b)}^n=\left({}_0^n\right)a^n{b}^0+\left({}_1^n\right)a^{n-1}{b}^1+\left({}_2^n\right)a^{n-2}{b}^2+...$$$$+\left({}_n^n\right)a^{n-n}{b}^n$$$$\Rightarrow a=b=1\Rightarrow A=B={(1+1)}^n=2^n$$

Commented by Matica last updated on 26/Dec/22

$$Thankyou!$$

Answered by mr W last updated on 26/Dec/22

$$f\left(x\right)={(1+x)}^n=C_0^n+C_1^{n}x+C_2^n{x}^2+...+C_n^n{x}^n\left(i\right)$$$$ontheothersideacc.taylorseries:$$$$f\left(x\right)=f\left(0\right)+f^{\prime }\left(0\right)x+\frac{f^{″}\left(0\right)}{2!}{x}^2+\frac{f^{‴}\left(0\right)}{3!}{x}^3+...+\frac{f^{\left(n\right)}\left(0\right)}{n!}{x}^n+\frac{f^{(n+1)}\left(0\right)}{(n+1)!}{x}^{n+1}+...\left(ii\right)$$$$compare\left(i\right)and\left(ii\right):$$$$\frac{f^{\left(k\right)}\left(0\right)}{k!}=C_k^{n}for0⩽k⩽nand$$$$\frac{f^{\left(k\right)}\left(0\right)}{k!}=0fork⩾n+1$$$$\Rightarrow f\left(0\right)+f^{\prime }\left(0\right)x+\frac{f^{″}\left(0\right)}{2!}{x}^2+\frac{f^{‴}\left(0\right)}{3!}{x}^3+...+\frac{f^{\left(n\right)}\left(0\right)}{n!}{x}^n={(1+x)}^n$$$$setx=1,$$$$\Rightarrow f\left(0\right)+f^{\prime }\left(0\right)+\frac{f^{″}\left(0\right)}{2!}+\frac{f^{‴}\left(0\right)}{3!}+...+\frac{f^{\left(n\right)}\left(0\right)}{n!}={(1+1)}^n=2^n✓$$

Commented by Matica last updated on 26/Dec/22

$$Thankyou!$$

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