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Find-x-1-3x-5x-2-7x-3-9x-4-11x-5-15-

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Question Number 210516 by hardmath last updated on 11/Aug/24
Find: x = ? 1 + 3x + 5x^2 + 7x^3 + 9x^4 + 11x^5 + ... = 15
$$\mathrm{Find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$$$\mathrm{1}\:+\:\mathrm{3x}\:+\:\mathrm{5x}^{\mathrm{2}} \:+\:\mathrm{7x}^{\mathrm{3}} \:+\:\mathrm{9x}^{\mathrm{4}} \:+\:\mathrm{11x}^{\mathrm{5}} \:+\:…\:=\:\mathrm{15} \\ $$
Answered by mm1342 last updated on 12/Aug/24
it is necessary −1<x<1 s=1+3x+5x^2 +7x^3 +9x^4 +... xs=x+3x^2 +5x^3 +7x^4 +9x^5 +... (1−x)s=−1+2(1+x+x^2 +x^3 +x^4 +...) (1−x)s=−1+(2/(1−x))=((x+1)/(1−x)) ⇒s=((x+1)/((x−1)^2 ))=15 ⇒ x>0 15x^2 −30x+15=x+1 15x^2 −31x+14=0 x=((31±11)/(30))⇒x=(7/5) × & x= ((20)/(30))=(2/3) ✓
$${it}\:{is}\:{necessary}\:\:\:\:−\mathrm{1}<{x}<\mathrm{1} \\ $$$${s}=\mathrm{1}+\mathrm{3}{x}+\mathrm{5}{x}^{\mathrm{2}} +\mathrm{7}{x}^{\mathrm{3}} +\mathrm{9}{x}^{\mathrm{4}} +… \\ $$$${xs}={x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}{x}^{\mathrm{3}} +\mathrm{7}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{5}} +… \\ $$$$\left(\mathrm{1}−{x}\right){s}=−\mathrm{1}+\mathrm{2}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +…\right) \\ $$$$\left(\mathrm{1}−{x}\right){s}=−\mathrm{1}+\frac{\mathrm{2}}{\mathrm{1}−{x}}=\frac{{x}+\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow{s}=\frac{{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{15}\:\:\:\Rightarrow\:{x}>\mathrm{0} \\ $$$$\mathrm{15}{x}^{\mathrm{2}} −\mathrm{30}{x}+\mathrm{15}={x}+\mathrm{1} \\ $$$$\mathrm{15}{x}^{\mathrm{2}} −\mathrm{31}{x}+\mathrm{14}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{31}\pm\mathrm{11}}{\mathrm{30}}\Rightarrow{x}=\frac{\mathrm{7}}{\mathrm{5}}\:× \\ $$$$\:\&\:\:{x}=\:\frac{\mathrm{20}}{\mathrm{30}}=\frac{\mathrm{2}}{\mathrm{3}}\:\checkmark \\ $$$$ \\ $$
Commented by Frix last updated on 11/Aug/24
Method ok but sign error: −1+(2/(1−x))=−((x+1)/(x−1))
$$\mathrm{Method}\:\mathrm{ok}\:\mathrm{but}\:\mathrm{sign}\:\mathrm{error}:\:−\mathrm{1}+\frac{\mathrm{2}}{\mathrm{1}−{x}}=−\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}} \\ $$
Commented by mm1342 last updated on 12/Aug/24
T
$$\:\underbrace{{T}} \\ $$
Answered by Frix last updated on 11/Aug/24
Σ_(k=1) ^n (2k−1)x^(k−1) = =(((2n−1)x^(n+1) −(2n+1)x^n +x+1)/((x−1)^2 ))=15 (2n−1)x^(n+1) −(2n+1)x^n −15x^2 +31x−14=0 −1<x<1: lim_(n→∞) ((2n−1)x^(n+1) −(2n+1)x^n ) =0 −15x^2 +31x−14=0 x=(2/3) Test: lim_(n→∞ ) (Σ_(k=1) ^n (2k−1)((2/3))^(k−1) ) = =lim_(n→∞) (15−3(2n+5)(2^n /3^n )) =15
$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}{k}−\mathrm{1}\right){x}^{{k}−\mathrm{1}} = \\ $$$$=\frac{\left(\mathrm{2}{n}−\mathrm{1}\right){x}^{{n}+\mathrm{1}} −\left(\mathrm{2}{n}+\mathrm{1}\right){x}^{{n}} +{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{15} \\ $$$$\left(\mathrm{2}{n}−\mathrm{1}\right){x}^{{n}+\mathrm{1}} −\left(\mathrm{2}{n}+\mathrm{1}\right){x}^{{n}} −\mathrm{15}{x}^{\mathrm{2}} +\mathrm{31}{x}−\mathrm{14}=\mathrm{0} \\ $$$$−\mathrm{1}<{x}<\mathrm{1}: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\left(\mathrm{2}{n}−\mathrm{1}\right){x}^{{n}+\mathrm{1}} −\left(\mathrm{2}{n}+\mathrm{1}\right){x}^{{n}} \right)\:=\mathrm{0} \\ $$$$−\mathrm{15}{x}^{\mathrm{2}} +\mathrm{31}{x}−\mathrm{14}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{Test}: \\ $$$$\underset{{n}\rightarrow\infty\:} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{k}−\mathrm{1}} \right)\:= \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{15}−\mathrm{3}\left(\mathrm{2}{n}+\mathrm{5}\right)\frac{\mathrm{2}^{{n}} }{\mathrm{3}^{{n}} }\right)\:=\mathrm{15} \\ $$
Commented by hardmath last updated on 11/Aug/24
thank you dear professors
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$
Answered by mr W last updated on 12/Aug/24
Σ_(k=0) ^∞ x^k =(1/(1−x)) for ∣x∣<1 Σ_(k=0) ^∞ kx^(k−1) =(1/((1−x)^2 )) Σ_(k=0) ^∞ 2kx^k =((2x)/((1−x)^2 )) Σ_(k=0) ^∞ (2k+1)x^k =((2x)/((1−x)^2 ))+(1/(1−x))=((1+x)/((1−x)^2 )) Σ_(k=0) ^∞ (2k+1)x^k =((1+x)/((1−x)^2 ))=15 ⇒15x^2 −31x+14=0 ⇒x=((31±11)/(30))=(7/5) (>1 rejected) or (2/3)
$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{k}} =\frac{\mathrm{1}}{\mathrm{1}−{x}}\:{for}\:\mid{x}\mid<\mathrm{1} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{kx}^{{k}−\mathrm{1}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}{kx}^{{k}} =\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{k}+\mathrm{1}\right){x}^{{k}} =\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−{x}}=\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{k}+\mathrm{1}\right){x}^{{k}} =\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\mathrm{15} \\ $$$$\Rightarrow\mathrm{15}{x}^{\mathrm{2}} −\mathrm{31}{x}+\mathrm{14}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{31}\pm\mathrm{11}}{\mathrm{30}}=\frac{\mathrm{7}}{\mathrm{5}}\:\left(>\mathrm{1}\:{rejected}\right)\:{or}\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by hardmath last updated on 12/Aug/24
thank you dear professors
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$

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