Question Number 141598 by Gbenga last updated on 20/May/21
Answered by TheSupreme last updated on 21/May/21
$$\frac{{A}}{{x}+\mathrm{3}}+\frac{{B}}{{x}+\mathrm{2}}=\frac{−\mathrm{2021}}{\left({x}+\mathrm{3}\right)\left({x}+\mathrm{2}\right)}=\frac{{Ax}+\mathrm{2}{A}+{Bx}+\mathrm{3}{B}}{\left({x}+\mathrm{3}\right)\left({x}+\mathrm{2}\right)} \\ $$$${A}+{B}=\mathrm{0} \\ $$$$\mathrm{2}{A}+\mathrm{3}{B}=−\mathrm{2021} \\ $$$${A}=\mathrm{2021} \\ $$$${B}=−\mathrm{2021} \\ $$$$\Sigma\frac{\mathrm{2021}}{\left({x}+\mathrm{3}\right)}−\frac{\mathrm{2021}}{\left({x}+\mathrm{2}\right)} \\ $$$${telescopic}\:{series} \\ $$$$\frac{\mathrm{2021}}{\mathrm{6}}−\frac{\mathrm{2021}}{\mathrm{5}}+\frac{\mathrm{2021}}{\mathrm{7}}−\frac{\mathrm{2021}}{\mathrm{6}}+\frac{\mathrm{2021}}{\mathrm{8}}−… \\ $$$$\frac{\mathrm{2021}}{\left(\mathrm{2018}+\mathrm{3}\right)}−\frac{\mathrm{2021}}{\mathrm{5}}=\frac{\mathrm{2021}}{\mathrm{2021}}−\frac{\mathrm{2021}}{\mathrm{5}}=\mathrm{1}−\mathrm{404}.\mathrm{2}=−\mathrm{403}.\mathrm{2} \\ $$
Commented by Gbenga last updated on 21/May/21
$$\boldsymbol{{T}}{hanks}!! \\ $$