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Question Number 81439 by jagoll last updated on 13/Feb/20

$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}$$$${\mathrm{f}}^{\left(100\right)}\left(0\right)=?$$

Commented by john santu last updated on 13/Feb/20

$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{\mathrm{n}+2}}{\mathrm{n}!}$$$$\Rightarrow \mathrm{n}+2=100\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{n}=0}{\sum ^{98}}\frac{{\mathrm{x}}^{100}}{98!}$$$$\mathrm{f}{}^{\left(100\right)}\left(0\right)=\frac{100!}{98!}=100\times 99=9900$$

Commented by abdomathmax last updated on 13/Feb/20

$$f^{\left(n\right)}\left(x\right)=\sum _{k=0}^n{C}_n^k{\left(x^2\right)}^{\left(k\right)}{\left(e^x\right)}^{(n-k)}$$$$=C_n^0{x}^2{e}^x+2x{C}_n^1{e}^x+2C_n^2{e}^x$$$$=x^2{e}^x+2nx{e}^x+2\times \frac{n!}{2!(n-2)!}{e}^x$$$$=(x^2+2nx+n(n-1))e^x\Rightarrow$$$$f^{\left(100\right)}\left(0\right)=100\times 99=9900$$

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