Question Number 216489 by manxsol last updated on 09/Feb/25
$${find}\:\:{residuo} \\ $$$$\:\:\:\:\:\:\:\:\frac{{x}^{\mathrm{21}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$
Commented by issac last updated on 09/Feb/25
$$\mathrm{Q216493} \\ $$
Answered by MrGaster last updated on 09/Feb/25
$$\mathrm{Let}\:{f}\left({x}\right)=\frac{{x}^{\mathrm{21}} +{x}^{\mathrm{2}} +\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\mathrm{Residue}\:\mathrm{at}\:{x}=\mathrm{1} \\ $$$$\mathrm{Let}\:{z}={x}\:{x}−\mathrm{1}\Rightarrow{x}={z}+\mathrm{1} \\ $$$${f}\left({z}+\mathrm{1}\right)=\frac{\left({z}+\mathrm{1}\right)^{\mathrm{21}} +\left({z}+\mathrm{1}\right)^{\mathrm{2}} +\left({z}+\mathrm{1}\right)+\mathrm{1}}{{z}^{\mathrm{3}} } \\ $$$$\left({z}+\mathrm{1}\right)^{\mathrm{21}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{21}} {\sum}}\begin{pmatrix}{\mathrm{21}}\\{{k}}\end{pmatrix}{z}^{{k}} \\ $$$$\left({z}+\mathrm{1}\right)^{\mathrm{2}} ={z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{1} \\ $$$$\left({z}+\mathrm{1}\right)={z}+\mathrm{1} \\ $$$${f}\left({z}+\mathrm{1}\right)=\frac{\underset{{k}=\mathrm{0}} {\overset{\mathrm{21}} {\sum}}\begin{pmatrix}{\mathrm{21}}\\{{k}}\end{pmatrix}{z}^{{k}} +{z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{1}+{z}+\mathrm{1}+\mathrm{1}}{{z}^{\mathrm{3}} } \\ $$$${f}\left({z}+\mathrm{1}\right)=\frac{\underset{{k}=\mathrm{0}} {\overset{\mathrm{21}} {\sum}}\begin{pmatrix}{\mathrm{21}}\\{{k}}\end{pmatrix}{z}^{{k}} +{z}^{\mathrm{2}} +\mathrm{3}{z}+\mathrm{3}}{{z}^{\mathrm{3}} } \\ $$$${z}^{\mathrm{2}} \mathrm{in}\left({z}+\mathrm{1}\right)^{\mathrm{21}} \mathrm{is}\begin{pmatrix}{\mathrm{21}}\\{\mathrm{2}}\end{pmatrix} \\ $$$${z}^{\mathrm{2}} \mathrm{in}\:{z}^{\mathrm{2}} +\mathrm{3}{z}+\mathrm{3}\:\mathrm{is}\:\mathrm{1} \\ $$$$\mathrm{Residue}=\begin{pmatrix}{\mathrm{21}}\\{\mathrm{2}}\end{pmatrix}+\mathrm{1}=\mathrm{210}+\mathrm{1}=\mathrm{211} \\ $$$$\mathrm{Residue}=\mathrm{211} \\ $$