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Question Number 141598 by Gbenga last updated on 20/May/21

Answered by TheSupreme last updated on 21/May/21

(A/(x+3))+(B/(x+2))=((−2021)/((x+3)(x+2)))=((Ax+2A+Bx+3B)/((x+3)(x+2))) A+B=0 2A+3B=−2021 A=2021 B=−2021 Σ((2021)/((x+3)))−((2021)/((x+2))) telescopic series ((2021)/6)−((2021)/5)+((2021)/7)−((2021)/6)+((2021)/8)−... ((2021)/((2018+3)))−((2021)/5)=((2021)/(2021))−((2021)/5)=1−404.2=−403.2

$$\frac{{A}}{{x}+\mathrm{3}}+\frac{{B}}{{x}+\mathrm{2}}=\frac{−\mathrm{2021}}{\left({x}+\mathrm{3}\right)\left({x}+\mathrm{2}\right)}=\frac{{Ax}+\mathrm{2}{A}+{Bx}+\mathrm{3}{B}}{\left({x}+\mathrm{3}\right)\left({x}+\mathrm{2}\right)} \\ $$$${A}+{B}=\mathrm{0} \\ $$$$\mathrm{2}{A}+\mathrm{3}{B}=−\mathrm{2021} \\ $$$${A}=\mathrm{2021} \\ $$$${B}=−\mathrm{2021} \\ $$$$\Sigma\frac{\mathrm{2021}}{\left({x}+\mathrm{3}\right)}−\frac{\mathrm{2021}}{\left({x}+\mathrm{2}\right)} \\ $$$${telescopic}\:{series} \\ $$$$\frac{\mathrm{2021}}{\mathrm{6}}−\frac{\mathrm{2021}}{\mathrm{5}}+\frac{\mathrm{2021}}{\mathrm{7}}−\frac{\mathrm{2021}}{\mathrm{6}}+\frac{\mathrm{2021}}{\mathrm{8}}−... \\ $$$$\frac{\mathrm{2021}}{\left(\mathrm{2018}+\mathrm{3}\right)}−\frac{\mathrm{2021}}{\mathrm{5}}=\frac{\mathrm{2021}}{\mathrm{2021}}−\frac{\mathrm{2021}}{\mathrm{5}}=\mathrm{1}−\mathrm{404}.\mathrm{2}=−\mathrm{403}.\mathrm{2} \\ $$

Commented by Gbenga last updated on 21/May/21

Thanks!!

$$\boldsymbol{{T}}{hanks}!! \\ $$

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