Question

If $\frac{ax+b}{(cx+d)x^{2020}}=\frac{A_1}{x}+\frac{A_2}{x^2}+......+\frac{A_{2020}}{x^{2020}}+\frac{B}{cx+d}$, then $A_{2019}=$

• A. $\frac{ad-bc}{d^3}$
• B. $\frac{-(ad-bc)}{d^2}$
• C. $\frac{-(ad-bc)}{d^3}$
• D. $\frac{(ad-bc)}{d^2}$

Answer 1

The correct option is D $\frac{(ad-bc)}{d^2}$

$\frac{ax+b}{(cx+d)x^{2020}}=\frac{A_1}{x}+\frac{A_2}{x^2}+\frac{A_3}{x^3}+..........+\frac{A_{2020}}{x^{2020}}+\frac{B}{cx+d}$

$⟹ax+b=A_1(cx+d)x^{2019}+A_2(cx+d)x^{2018}+.....+A_{2018}(cx+d)x^2+A_{2019}(cx+d)x+A_{2020}(cx+d)+B{x}^{2020}$

$⟹ax+b=(d{A}_{2019}+c{A}_{2020})x+d{A}_{2020}$

$⟹a=d{A}_{2019}+c{A}_{2020}$ and $b=d{A}_{2020}$

$⟹a=d{A}_{2019}+\frac{bc}{d}$ and $A_{2020}=\frac{b}{d}$

$⟹A_{2019}=\frac{a}{d}-\frac{bc}{d^2}$

$\therefore A_{2019}=\frac{ad-bc}{d^2}$ (Ans)

Option D is correct.