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Question Number 55869 by mr W last updated on 05/Mar/19

if a_(n+2) =(a_(n+1) ^3 /a_n ^2 ) and a_1 =2, a_2 =4 find a_n =?

$${if}\:\boldsymbol{{a}}_{\boldsymbol{{n}}+\mathrm{2}} =\frac{\boldsymbol{{a}}_{\boldsymbol{{n}}+\mathrm{1}} ^{\mathrm{3}} }{\boldsymbol{{a}}_{\boldsymbol{{n}}} ^{\mathrm{2}} }\:{and}\:{a}_{\mathrm{1}} =\mathrm{2},\:{a}_{\mathrm{2}} =\mathrm{4} \\ $$$${find}\:\boldsymbol{{a}}_{\boldsymbol{{n}}} =? \\ $$

Commented by Tawa1 last updated on 05/Mar/19

sir, please check question 55785 you solved. I asked something ..

$$\mathrm{sir},\:\mathrm{please}\:\mathrm{check}\:\mathrm{question}\:\:\mathrm{55785}\:\:\mathrm{you}\:\mathrm{solved}.\:\mathrm{I}\:\mathrm{asked}\:\mathrm{something}\:.. \\ $$

Commented by MJS last updated on 05/Mar/19

a_n =(a_2 ^((2^(n−1) −1)) /a_1 ^((2^(n−1) −2)) )

$${a}_{{n}} =\frac{{a}_{\mathrm{2}} ^{\left(\mathrm{2}^{{n}−\mathrm{1}} −\mathrm{1}\right)} }{{a}_{\mathrm{1}} ^{\left(\mathrm{2}^{{n}−\mathrm{1}} −\mathrm{2}\right)} } \\ $$

Answered by MJS last updated on 05/Mar/19

a_n =2^((2^(n−1) )) =2^(2^n /2)

$${a}_{{n}} =\mathrm{2}^{\left(\mathrm{2}^{{n}−\mathrm{1}} \right)} =\mathrm{2}^{\frac{\mathrm{2}^{{n}} }{\mathrm{2}}} \\ $$

Commented by mr W last updated on 05/Mar/19

thanks sir!

$${thanks}\:{sir}! \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19

(a_(n+2) /a_(n+1) )=((a_(n+1) /a_n ))^2 (a_3 /a_2 )=((a_2 /a_1 ))^2 =((4/2))^2 =4 (a_2 /a_1 )=2=2^1 (a_3 /a_2 )=4=2^2 (a_4 /a_3 )=((a_3 /a_2 ))^2 =4^2 =2^4 (a_5 /a_4 )=((a_4 /a_3 ))^2 =2^8 .... a_1 =a_1 =2→2^2^0 a_2 =2^1 a_1 =4=2^2 →2^2^1 a_3 =2^2 a_2 =2^2 ×2^2 =2^4 →2^2^2 a_4 =2^4 a_3 =2^4 ×2^4 =2^8 →2^2^3 a_5 =2^8 ×a^4 =2^8 ×2^8 →2^2^4 so a_n =2^2^(n−1)

$$\frac{{a}_{{n}+\mathrm{2}} }{{a}_{{n}+\mathrm{1}} }=\left(\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }\right)^{\mathrm{2}} \\ $$$$\frac{{a}_{\mathrm{3}} }{{a}_{\mathrm{2}} }=\left(\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }\right)^{\mathrm{2}} =\left(\frac{\mathrm{4}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }=\mathrm{2}=\mathrm{2}^{\mathrm{1}} \\ $$$$\frac{{a}_{\mathrm{3}} }{{a}_{\mathrm{2}} }=\mathrm{4}=\mathrm{2}^{\mathrm{2}} \\ $$$$\frac{{a}_{\mathrm{4}} }{{a}_{\mathrm{3}} }=\left(\frac{{a}_{\mathrm{3}} }{{a}_{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} =\mathrm{2}^{\mathrm{4}} \\ $$$$\frac{{a}_{\mathrm{5}} }{{a}_{\mathrm{4}} }=\left(\frac{{a}_{\mathrm{4}} }{{a}_{\mathrm{3}} }\right)^{\mathrm{2}} =\mathrm{2}^{\mathrm{8}} \\ $$$$.... \\ $$$${a}_{\mathrm{1}} ={a}_{\mathrm{1}} =\mathrm{2}\rightarrow\mathrm{2}^{\mathrm{2}^{\mathrm{0}} } \\ $$$${a}_{\mathrm{2}} =\mathrm{2}^{\mathrm{1}} {a}_{\mathrm{1}} =\mathrm{4}=\mathrm{2}^{\mathrm{2}} \rightarrow\mathrm{2}^{\mathrm{2}^{\mathrm{1}} } \\ $$$${a}_{\mathrm{3}} =\mathrm{2}^{\mathrm{2}} {a}_{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{2}} =\mathrm{2}^{\mathrm{4}} \rightarrow\mathrm{2}^{\mathrm{2}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{4}} =\mathrm{2}^{\mathrm{4}} {a}_{\mathrm{3}} =\mathrm{2}^{\mathrm{4}} ×\mathrm{2}^{\mathrm{4}} =\mathrm{2}^{\mathrm{8}} \rightarrow\mathrm{2}^{\mathrm{2}^{\mathrm{3}} } \\ $$$${a}_{\mathrm{5}} =\mathrm{2}^{\mathrm{8}} ×{a}^{\mathrm{4}} =\mathrm{2}^{\mathrm{8}} ×\mathrm{2}^{\mathrm{8}} \rightarrow\mathrm{2}^{\mathrm{2}^{\mathrm{4}} } \\ $$$${so}\:{a}_{{n}} =\mathrm{2}^{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$

Commented by mr W last updated on 05/Mar/19

very nice sir! thank you!

$${very}\:{nice}\:{sir}!\:{thank}\:{you}! \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19

thank you sir brain cell gets charged and gets activated to solve these type of problems...

$${thank}\:{you}\:{sir}\:{brain}\:{cell}\:{gets}\:{charged}\:{and}\:{gets}\:{activated} \\ $$$${to}\:{solve}\:{these}\:{type}\:{of}\:{problems}... \\ $$

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