Question Number 22503 by Tinkutara last updated on 19/Oct/17
$$\mathrm{If}\:\left({a}\:+\:{bx}\right)^{−\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{4}}\:−\:\mathrm{3}{x}\:+\:…,\:\mathrm{then}\:\left({a},\:{b}\right)\:= \\ $$
Commented by mrW1 last updated on 19/Oct/17
![(a+bx)^(−2) =a^(−2) (1+((bx)/a))^(−2) =a^(−2) [1−2×((bx)/a)+...] a^(−2) =(1/4)=2^(−2) ⇒a=2 −2a^(−2) ×(b/a)=−3 ⇒b=3×a^3 /2=3×2^3 /2=12](https://www.tinkutara.com/question/Q22504.png)
$$\left(\mathrm{a}+\mathrm{bx}\right)^{−\mathrm{2}} =\mathrm{a}^{−\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{bx}}{\mathrm{a}}\right)^{−\mathrm{2}} =\mathrm{a}^{−\mathrm{2}} \left[\mathrm{1}−\mathrm{2}×\frac{\mathrm{bx}}{\mathrm{a}}+…\right] \\ $$$$\mathrm{a}^{−\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{2}^{−\mathrm{2}} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{2} \\ $$$$−\mathrm{2a}^{−\mathrm{2}} ×\frac{\mathrm{b}}{\mathrm{a}}=−\mathrm{3} \\ $$$$\Rightarrow\mathrm{b}=\mathrm{3}×\mathrm{a}^{\mathrm{3}} /\mathrm{2}=\mathrm{3}×\mathrm{2}^{\mathrm{3}} /\mathrm{2}=\mathrm{12} \\ $$
Commented by Tinkutara last updated on 19/Oct/17
$$\left(\mathrm{2},\mathrm{12}\right)\:\mathrm{wrong}. \\ $$
Commented by mrW1 last updated on 19/Oct/17
$$\mathrm{a}=\mathrm{2},\mathrm{b}=\mathrm{12} \\ $$
Commented by ajfour last updated on 19/Oct/17
$${correct}\:{answer}\:{please}. \\ $$
Answered by ajfour last updated on 19/Oct/17
$$\left(\mathrm{1}+{x}\right)^{{n}} =\mathrm{1}+\frac{{n}}{\mathrm{1}!}{x}+\frac{{n}\left({n}−\mathrm{2}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +… \\ $$$${coeff}.\:{of}\:{x}^{\mathrm{0}} \:{in}\:\frac{\left(\mathrm{1}+\frac{{bx}}{{a}}\right)^{−\mathrm{2}} }{{a}^{\mathrm{2}} }\:\:\:{is} \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\Rightarrow\:\:{a}=\pm\mathrm{2} \\ $$$${coeff}.\:{of}\:{x}\:{in}\:\frac{\left(\mathrm{1}+\frac{{bx}}{{a}}\right)^{−\mathrm{2}} }{{a}^{\mathrm{2}} }\:{is} \\ $$$$=\frac{−\mathrm{2}{b}}{{a}^{\mathrm{3}} }\:=\:−\mathrm{3} \\ $$$$\Rightarrow\:\:\:{b}=\pm\mathrm{12}\:.\:\:\: \\ $$$$\left({a},\:{b}\right)\:{is}\:\left(\mathrm{2},\:\mathrm{12}\right)\:\:{or}\:\left(−\mathrm{2},−\mathrm{12}\right)\:. \\ $$
Commented by Tinkutara last updated on 19/Oct/17
$$\mathrm{But}\:\mathrm{answer}\:\mathrm{given}\:\mathrm{is}\:\mathrm{only}\:\left(−\mathrm{2},−\mathrm{12}\right). \\ $$$$\mathrm{Why}? \\ $$