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Question Number 22547 by Tinkutara last updated on 20/Oct/17

If α = (5/(2!3)) + ((5.7)/(3!3^2 )) + ((5.7.9)/(4!3^3 )) ,... then find the value of α^2 + 4α.

$$\mathrm{If}\:\alpha\:=\:\frac{\mathrm{5}}{\mathrm{2}!\mathrm{3}}\:+\:\frac{\mathrm{5}.\mathrm{7}}{\mathrm{3}!\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{5}.\mathrm{7}.\mathrm{9}}{\mathrm{4}!\mathrm{3}^{\mathrm{3}} }\:,...\:\mathrm{then}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\alpha^{\mathrm{2}} \:+\:\mathrm{4}\alpha. \\ $$

Answered by ajfour last updated on 20/Oct/17

(1+x)^n =1+nx+((n(n−1))/(2!))x^2 +... let x=−(2/3) ,n→−n (1+x)^(−n) ∣_(x=−(2/3)) =1+((2n)/3)+((2n(2n+2))/(2!3^2 )) +((2n(2n+2)(2n+4))/(3!3^3 ))+.. let n=3/2 ⇒(1−(2/3))^(−3/2) =1+1+(5/(2!3))+((5.7)/(3!3^2 ))+.. ⇒ 3(√3) =2+α α=3(√3)−2 α^2 +4α=α(α+4) =(3(√3)−2)(3(√3)+2) =23 .

$$\left(\mathrm{1}+{x}\right)^{{n}} =\mathrm{1}+{nx}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +... \\ $$$${let}\:{x}=−\frac{\mathrm{2}}{\mathrm{3}}\:\:,{n}\rightarrow−{n} \\ $$$$\left(\mathrm{1}+{x}\right)^{−{n}} \mid_{{x}=−\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{1}+\frac{\mathrm{2}{n}}{\mathrm{3}}+\frac{\mathrm{2}{n}\left(\mathrm{2}{n}+\mathrm{2}\right)}{\mathrm{2}!\mathrm{3}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{2}{n}\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{4}\right)}{\mathrm{3}!\mathrm{3}^{\mathrm{3}} }+.. \\ $$$${let}\:\boldsymbol{{n}}=\mathrm{3}/\mathrm{2} \\ $$$$\Rightarrow\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\right)^{−\mathrm{3}/\mathrm{2}} =\mathrm{1}+\mathrm{1}+\frac{\mathrm{5}}{\mathrm{2}!\mathrm{3}}+\frac{\mathrm{5}.\mathrm{7}}{\mathrm{3}!\mathrm{3}^{\mathrm{2}} }+.. \\ $$$$\Rightarrow\:\mathrm{3}\sqrt{\mathrm{3}}\:=\mathrm{2}+\alpha \\ $$$$\:\:\alpha=\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{2} \\ $$$$\:\:\alpha^{\mathrm{2}} +\mathrm{4}\alpha=\alpha\left(\alpha+\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:=\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{2}\right)\left(\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:=\mathrm{23}\:. \\ $$$$ \\ $$

Commented by ajfour last updated on 20/Oct/17

seeking 𝛂 .

$${seeking}\:\boldsymbol{\alpha}\:. \\ $$

Commented by Tinkutara last updated on 20/Oct/17

How you have written x=−(2/3), n=(3/2)?

$$\mathrm{How}\:\mathrm{you}\:\mathrm{have}\:\mathrm{written}\:{x}=−\frac{\mathrm{2}}{\mathrm{3}},\:{n}=\frac{\mathrm{3}}{\mathrm{2}}? \\ $$

Commented by Tinkutara last updated on 03/Jan/18

Thanks Sir!

Commented by ajfour last updated on 03/Jan/18

α need to be generated, 3, 3^2 , 3^3 ,.. in denominator coaxed me to put x=(2/3) , ...

$$\alpha\:{need}\:{to}\:{be}\:{generated},\:\mathrm{3},\:\mathrm{3}^{\mathrm{2}} ,\:\mathrm{3}^{\mathrm{3}} ,.. \\ $$$${in}\:{denominator}\:{coaxed}\:{me}\:{to} \\ $$$${put}\:{x}=\frac{\mathrm{2}}{\mathrm{3}}\:,\:... \\ $$

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