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Question-157576

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Question Number 157576 by amin96 last updated on 24/Oct/21
Commented by quvonch3737 last updated on 24/Oct/21
1+(1/x)+(9/x^2 )+((16)/x^3 )+...=S /×(1/x) (1/x)+(4/x^2 )+(9/x^3 )+((16)/x^4 )+...=(S/x) S−(S/x)=1+(3/x)+(5/x^2 )+(7/x^3 )+(9/x^4 )+...=91 S(1−(1/x))=91 S=((91x)/(x−1r))
$$\mathrm{1}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{9}}{{x}^{\mathrm{2}} }+\frac{\mathrm{16}}{{x}^{\mathrm{3}} }+…={S}\:\:/×\frac{\mathrm{1}}{{x}} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }+\frac{\mathrm{9}}{{x}^{\mathrm{3}} }+\frac{\mathrm{16}}{{x}^{\mathrm{4}} }+…=\frac{{S}}{{x}} \\ $$$$\:{S}−\frac{{S}}{{x}}=\mathrm{1}+\frac{\mathrm{3}}{{x}}+\frac{\mathrm{5}}{{x}^{\mathrm{2}} }+\frac{\mathrm{7}}{{x}^{\mathrm{3}} }+\frac{\mathrm{9}}{{x}^{\mathrm{4}} }+…=\mathrm{91} \\ $$$${S}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=\mathrm{91} \\ $$$${S}=\frac{\mathrm{91}{x}}{{x}−\mathrm{1}{r}} \\ $$
Commented by Rasheed.Sindhi last updated on 25/Oct/21
Slightly defferent way... 1+(3/x)+(5/x^2 )+(7/x^3 )+(9/x^4 )+...=91 1+(4/x)+(9/x^2 )+((16)/x^3 )+((25)/x^4 )+...=S S−91=(1/x)+(4/x^2 )+(9/x^3 )+((16)/x^4 )+... =(1/x)(1+(4/x)+(9/x^2 )+((16)/x^3 )+...) =(1/x)(S) S−(S/x)=91 S(((x−1)/x))=91 S=((91x)/(x−1))
$${Slightly}\:\:{defferent}\:{way}… \\ $$$$\mathrm{1}+\frac{\mathrm{3}}{{x}}+\frac{\mathrm{5}}{{x}^{\mathrm{2}} }+\frac{\mathrm{7}}{{x}^{\mathrm{3}} }+\frac{\mathrm{9}}{{x}^{\mathrm{4}} }+…=\mathrm{91} \\ $$$$\mathrm{1}+\frac{\mathrm{4}}{{x}}+\frac{\mathrm{9}}{{x}^{\mathrm{2}} }+\frac{\mathrm{16}}{{x}^{\mathrm{3}} }+\frac{\mathrm{25}}{{x}^{\mathrm{4}} }+…={S} \\ $$$${S}−\mathrm{91}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }+\frac{\mathrm{9}}{{x}^{\mathrm{3}} }+\frac{\mathrm{16}}{{x}^{\mathrm{4}} }+… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{x}}\left(\mathrm{1}+\frac{\mathrm{4}}{{x}}+\frac{\mathrm{9}}{{x}^{\mathrm{2}} }+\frac{\mathrm{16}}{{x}^{\mathrm{3}} }+…\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{x}}\left({S}\right) \\ $$$${S}−\frac{{S}}{{x}}=\mathrm{91} \\ $$$${S}\left(\frac{{x}−\mathrm{1}}{{x}}\right)=\mathrm{91} \\ $$$${S}=\frac{\mathrm{91}{x}}{{x}−\mathrm{1}} \\ $$$$ \\ $$
Answered by mr W last updated on 25/Oct/21
let t=(1/x) S_1 =1+3t+5t^2 +7t^3 +9t^4 +...=91 tS_1 = t+3t^2 +5t^3 +7t^4 +9t^5 +... (1−t)S_1 =1+2(t+t^2 +t^3 +t^4 +...) (1−t)S_1 =−1+(2/(1−t)) (only if ∣t∣<1, i.e. ∣x∣>1) (2/((1−t)^2 ))−(1/(1−t))−S_1 =0 ⇒(1/(1−t))=((1±(√(1+8S_1 )))/4) =((1±(√(1+8×91)))/4)=((1±27)/4)=7,−((13)/2) ∣t∣=∣(1/x)∣<1 −1<t<1 ⇒0<1−t<2 ⇒(1/(1−t))>(1/2) ⇒only (1/(1−t))=7 is valid, −((13)/2) is rejected i.e. x=(1/t)=(7/6) S_2 =1+(4/x)+(9/x^2 )+((16)/x^3 )+((25)/x^4 )+... S_2 =1+4t+9t^2 +16t^3 +25t^4 + S_1 =1+3t+5t^2 +7t^3 +9t^4 +... S_2 −S_1 =t+4t^2 +9t^3 +16t^4 ((S_2 −S_1 )/t)=1+4t+9t^2 +16t^3 +... ((S_2 −S_1 )/t)=S_2 S_2 =(S_1 /(1−t))=91×7=637 ⇒1+(4/x)+(9/x^2 )+((16)/x^3 )+((25)/x^4 )+...=637
$${let}\:{t}=\frac{\mathrm{1}}{{x}} \\ $$$${S}_{\mathrm{1}} =\mathrm{1}+\mathrm{3}{t}+\mathrm{5}{t}^{\mathrm{2}} +\mathrm{7}{t}^{\mathrm{3}} +\mathrm{9}{t}^{\mathrm{4}} +…=\mathrm{91} \\ $$$${tS}_{\mathrm{1}} =\:{t}+\mathrm{3}{t}^{\mathrm{2}} +\mathrm{5}{t}^{\mathrm{3}} +\mathrm{7}{t}^{\mathrm{4}} +\mathrm{9}{t}^{\mathrm{5}} +… \\ $$$$\left(\mathrm{1}−{t}\right){S}_{\mathrm{1}} =\mathrm{1}+\mathrm{2}\left({t}+{t}^{\mathrm{2}} +{t}^{\mathrm{3}} +{t}^{\mathrm{4}} +…\right) \\ $$$$\left(\mathrm{1}−{t}\right){S}_{\mathrm{1}} =−\mathrm{1}+\frac{\mathrm{2}}{\mathrm{1}−{t}}\:\:\:\:\:\:\:\:\:\:\:\left({only}\:{if}\:\mid{t}\mid<\mathrm{1},\:{i}.{e}.\:\mid{x}\mid>\mathrm{1}\right) \\ $$$$\frac{\mathrm{2}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}−{t}}−{S}_{\mathrm{1}} =\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}−{t}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{8}{S}_{\mathrm{1}} }}{\mathrm{4}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{8}×\mathrm{91}}}{\mathrm{4}}=\frac{\mathrm{1}\pm\mathrm{27}}{\mathrm{4}}=\mathrm{7},−\frac{\mathrm{13}}{\mathrm{2}} \\ $$$$\mid{t}\mid=\mid\frac{\mathrm{1}}{{x}}\mid<\mathrm{1} \\ $$$$−\mathrm{1}<{t}<\mathrm{1} \\ $$$$\Rightarrow\mathrm{0}<\mathrm{1}−{t}<\mathrm{2} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}−{t}}>\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{only}\:\frac{\mathrm{1}}{\mathrm{1}−{t}}=\mathrm{7}\:{is}\:{valid},\:−\frac{\mathrm{13}}{\mathrm{2}}\:{is}\:{rejected} \\ $$$${i}.{e}.\:{x}=\frac{\mathrm{1}}{{t}}=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$ \\ $$$${S}_{\mathrm{2}} =\mathrm{1}+\frac{\mathrm{4}}{{x}}+\frac{\mathrm{9}}{{x}^{\mathrm{2}} }+\frac{\mathrm{16}}{{x}^{\mathrm{3}} }+\frac{\mathrm{25}}{{x}^{\mathrm{4}} }+… \\ $$$${S}_{\mathrm{2}} =\mathrm{1}+\mathrm{4}{t}+\mathrm{9}{t}^{\mathrm{2}} +\mathrm{16}{t}^{\mathrm{3}} +\mathrm{25}{t}^{\mathrm{4}} + \\ $$$${S}_{\mathrm{1}} =\mathrm{1}+\mathrm{3}{t}+\mathrm{5}{t}^{\mathrm{2}} +\mathrm{7}{t}^{\mathrm{3}} +\mathrm{9}{t}^{\mathrm{4}} +… \\ $$$${S}_{\mathrm{2}} −{S}_{\mathrm{1}} ={t}+\mathrm{4}{t}^{\mathrm{2}} +\mathrm{9}{t}^{\mathrm{3}} +\mathrm{16}{t}^{\mathrm{4}} \\ $$$$\frac{{S}_{\mathrm{2}} −{S}_{\mathrm{1}} }{{t}}=\mathrm{1}+\mathrm{4}{t}+\mathrm{9}{t}^{\mathrm{2}} +\mathrm{16}{t}^{\mathrm{3}} +… \\ $$$$\frac{{S}_{\mathrm{2}} −{S}_{\mathrm{1}} }{{t}}={S}_{\mathrm{2}} \\ $$$${S}_{\mathrm{2}} =\frac{{S}_{\mathrm{1}} }{\mathrm{1}−{t}}=\mathrm{91}×\mathrm{7}=\mathrm{637} \\ $$$$\Rightarrow\mathrm{1}+\frac{\mathrm{4}}{{x}}+\frac{\mathrm{9}}{{x}^{\mathrm{2}} }+\frac{\mathrm{16}}{{x}^{\mathrm{3}} }+\frac{\mathrm{25}}{{x}^{\mathrm{4}} }+…=\mathrm{637} \\ $$

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