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Question Number 216489 by manxsol last updated on 09/Feb/25

$$findresiduo$$$$\frac{x^{21}+x^2+x+1}{{(x-1)}^3}$$

Commented by issac last updated on 09/Feb/25

$$\mathrm{Q}216493$$

Answered by MrGaster last updated on 09/Feb/25

$$\mathrm{Let}f\left(x\right)=\frac{x^{21}+x^2+1}{{(x-1)}^3}$$$$\mathrm{Residue}\mathrm{at}x=1$$$$\mathrm{Let}z=xx-1\Rightarrow x=z+1$$$$f(z+1)=\frac{{(z+1)}^{21}+{(z+1)}^2+(z+1)+1}{z^3}$$$${(z+1)}^{21}=\underset{k=0}{\sum ^{21}}\left(\begin{array}{c}21\\ k\end{array}\right)z^k$$$${(z+1)}^2=z^2+2z+1$$$$(z+1)=z+1$$$$f(z+1)=\frac{\underset{k=0}{\sum ^{21}}\left(\begin{array}{c}21\\ k\end{array}\right)z^k+z^2+2z+1+z+1+1}{z^3}$$$$f(z+1)=\frac{\underset{k=0}{\sum ^{21}}\left(\begin{array}{c}21\\ k\end{array}\right)z^k+z^2+3z+3}{z^3}$$$$z^2\mathrm{in}{(z+1)}^{21}\mathrm{is}\left(\begin{array}{c}21\\ 2\end{array}\right)$$$$z^2\mathrm{in}{z}^2+3z+3\mathrm{is}1$$$$\mathrm{Residue}=\left(\begin{array}{c}21\\ 2\end{array}\right)+1=210+1=211$$$$\mathrm{Residue}=211$$

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