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Question Number 22503 by Tinkutara last updated on 19/Oct/17

If (a + bx)^(−2)  = (1/4) − 3x + ..., then (a, b) =

$$\mathrm{If}\:\left({a}\:+\:{bx}\right)^{−\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{4}}\:−\:\mathrm{3}{x}\:+\:...,\:\mathrm{then}\:\left({a},\:{b}\right)\:= \\ $$

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

$$\left(\mathrm{a}+\mathrm{bx}\right)^{−\mathrm{2}} =\mathrm{a}^{−\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{bx}}{\mathrm{a}}\right)^{−\mathrm{2}} =\mathrm{a}^{−\mathrm{2}} \left[\mathrm{1}−\mathrm{2}×\frac{\mathrm{bx}}{\mathrm{a}}+...\right] \\ $$$$\mathrm{a}^{−\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{2}^{−\mathrm{2}} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{2} \\ $$$$−\mathrm{2a}^{−\mathrm{2}} ×\frac{\mathrm{b}}{\mathrm{a}}=−\mathrm{3} \\ $$$$\Rightarrow\mathrm{b}=\mathrm{3}×\mathrm{a}^{\mathrm{3}} /\mathrm{2}=\mathrm{3}×\mathrm{2}^{\mathrm{3}} /\mathrm{2}=\mathrm{12} \\ $$

Commented by Tinkutara last updated on 19/Oct/17

(2,12) wrong.

$$\left(\mathrm{2},\mathrm{12}\right)\:\mathrm{wrong}. \\ $$

Commented by mrW1 last updated on 19/Oct/17

a=2,b=12

$$\mathrm{a}=\mathrm{2},\mathrm{b}=\mathrm{12} \\ $$

Commented by ajfour last updated on 19/Oct/17

correct answer please.

$${correct}\:{answer}\:{please}. \\ $$

Answered by ajfour last updated on 19/Oct/17

(1+x)^n =1+(n/(1!))x+((n(n−2))/(2!))x^2 +...  coeff. of x^0  in (((1+((bx)/a))^(−2) )/a^2 )   is  =(1/a^2 ) =(1/4)   ⇒  a=±2  coeff. of x in (((1+((bx)/a))^(−2) )/a^2 ) is  =((−2b)/a^3 ) = −3  ⇒   b=±12 .     (a, b) is (2, 12)  or (−2,−12) .

$$\left(\mathrm{1}+{x}\right)^{{n}} =\mathrm{1}+\frac{{n}}{\mathrm{1}!}{x}+\frac{{n}\left({n}−\mathrm{2}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +... \\ $$$${coeff}.\:{of}\:{x}^{\mathrm{0}} \:{in}\:\frac{\left(\mathrm{1}+\frac{{bx}}{{a}}\right)^{−\mathrm{2}} }{{a}^{\mathrm{2}} }\:\:\:{is} \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\Rightarrow\:\:{a}=\pm\mathrm{2} \\ $$$${coeff}.\:{of}\:{x}\:{in}\:\frac{\left(\mathrm{1}+\frac{{bx}}{{a}}\right)^{−\mathrm{2}} }{{a}^{\mathrm{2}} }\:{is} \\ $$$$=\frac{−\mathrm{2}{b}}{{a}^{\mathrm{3}} }\:=\:−\mathrm{3} \\ $$$$\Rightarrow\:\:\:{b}=\pm\mathrm{12}\:.\:\:\: \\ $$$$\left({a},\:{b}\right)\:{is}\:\left(\mathrm{2},\:\mathrm{12}\right)\:\:{or}\:\left(−\mathrm{2},−\mathrm{12}\right)\:. \\ $$

Commented by Tinkutara last updated on 19/Oct/17

But answer given is only (−2,−12).  Why?

$$\mathrm{But}\:\mathrm{answer}\:\mathrm{given}\:\mathrm{is}\:\mathrm{only}\:\left(−\mathrm{2},−\mathrm{12}\right). \\ $$$$\mathrm{Why}? \\ $$

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