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Question Number 130593 by mr W last updated on 27/Jan/21

if a_0 =1, a_1 =2 and a_(n+1) =(√(a_n a_(n−1) )) find a_n in terms of n.

$${if}\:{a}_{\mathrm{0}} =\mathrm{1},\:{a}_{\mathrm{1}} =\mathrm{2}\:{and}\:{a}_{{n}+\mathrm{1}} =\sqrt{{a}_{{n}} {a}_{{n}−\mathrm{1}} } \\ $$$${find}\:{a}_{{n}} \:{in}\:{terms}\:{of}\:{n}. \\ $$

Commented by malwan last updated on 27/Jan/21

a_n =2^(((((((2^n +(−1)^(n−1) )/3)))/2^(n−1) ))) n≥3 edited

$${a}_{{n}} \:=\mathrm{2}^{\left(\frac{\left(\frac{\mathrm{2}^{{n}} +\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{3}}\right)}{\mathrm{2}^{{n}−\mathrm{1}} }\right)} \\ $$$${n}\geqslant\mathrm{3}\:{edited} \\ $$

Commented by malwan last updated on 27/Jan/21

a_2 = (2×1)^(1/2) = 2^(1/2) a_3 = (2^(1/2) ×2)^(1/2) = 2^(3/4) a_4 = (2^(3/4) ×2^(1/2) )^(1/2) = 2^(5/8) a_5 = (2^(5/8) ×2^(3/4) )^(1/2) = 2^((11)/(16)) =2^(((((32+1)/3)))/2^(4 ) ) = 2^(((((2^5 +1^(5−1) )/3)))/2^(5−1) ) a_6 = (2^((11)/(16)) ×2^(5/8) )^(1/2) = 2^((21)/(32)) =2^(((((64−1)/3)))/2^5 ) =2^(((((2^6 +(−1)^(6−1) )/3)))/2^(6−1) ) a_7 = (2^((21)/(32)) ×2^((11)/(16)) )^(1/2) = 2^((43)/(64)) =2^(((((2^7 +(−1)^(7−1) )/3)))/2^(7−1) ) ∴ a_n = 2^(((((2^n +(−1)^(n−1) )/3)))/2^(n−1) )

$${a}_{\mathrm{2}} \:=\:\left(\mathrm{2}×\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${a}_{\mathrm{3}} \:=\:\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$${a}_{\mathrm{4}} \:=\:\left(\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} ×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\mathrm{2}^{\frac{\mathrm{5}}{\mathrm{8}}} \\ $$$${a}_{\mathrm{5}} \:=\:\left(\mathrm{2}^{\frac{\mathrm{5}}{\mathrm{8}}} ×\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\:\mathrm{2}^{\frac{\mathrm{11}}{\mathrm{16}}} \:=\mathrm{2}^{\frac{\left(\frac{\mathrm{32}+\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{2}^{\mathrm{4}\:} }} \\ $$$$=\:\mathrm{2}^{\frac{\left(\frac{\mathrm{2}^{\mathrm{5}} +\mathrm{1}^{\mathrm{5}−\mathrm{1}} }{\mathrm{3}}\right)}{\mathrm{2}^{\mathrm{5}−\mathrm{1}} }} \\ $$$${a}_{\mathrm{6}} \:=\:\left(\mathrm{2}^{\frac{\mathrm{11}}{\mathrm{16}}} ×\mathrm{2}^{\frac{\mathrm{5}}{\mathrm{8}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\:\mathrm{2}^{\frac{\mathrm{21}}{\mathrm{32}}} =\mathrm{2}^{\frac{\left(\frac{\mathrm{64}−\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{2}^{\mathrm{5}} }} \\ $$$$=\mathrm{2}^{\frac{\left(\frac{\mathrm{2}^{\mathrm{6}} +\left(−\mathrm{1}\right)^{\mathrm{6}−\mathrm{1}} }{\mathrm{3}}\right)}{\mathrm{2}^{\mathrm{6}−\mathrm{1}} }} \\ $$$${a}_{\mathrm{7}} \:=\:\left(\mathrm{2}^{\frac{\mathrm{21}}{\mathrm{32}}} ×\mathrm{2}^{\frac{\mathrm{11}}{\mathrm{16}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\:\mathrm{2}^{\frac{\mathrm{43}}{\mathrm{64}}} =\mathrm{2}^{\frac{\left(\frac{\mathrm{2}^{\mathrm{7}} +\left(−\mathrm{1}\right)^{\mathrm{7}−\mathrm{1}} }{\mathrm{3}}\right)}{\mathrm{2}^{\mathrm{7}−\mathrm{1}} }} \\ $$$$\therefore\:{a}_{{n}} \:=\:\mathrm{2}^{\frac{\left(\frac{\mathrm{2}^{{n}} +\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{3}}\right)}{\mathrm{2}^{{n}−\mathrm{1}} }} \\ $$

Commented by mr W last updated on 27/Jan/21

thank you! good!

$${thank}\:{you}!\:{good}! \\ $$

Answered by mindispower last updated on 27/Jan/21

⇒ln(a_(n+1) )−(1/2)ln(a_n )−(1/z)ln(a_(n−1) )=0 ln(a_n ).w_n ⇒w_(n+1) −(w_n /2)−(w_(n−1) /2)=0 X^2 −(X/2)−(1/2).0 X_1 =1,X_2 =−(1/2) w_n =a+b(−(1/2))^n w_0 =0⇒a+b=0,2a−b=2ln(2) a=(2/3)ln(2),b=−(2/3)ln(2) w_n =ln(2^(2/3) )−(2/3)ln(2)(−(1/2))^n =ln(2^(2/3) .2^(((−2)/3).(((−1)^n )/2^n )) ) =ln(2^((2^(n+1) −2(−1)^n )/(3.2^n )) )=w_n ,a_n =e^w_n =2^((2^(n+1) −2(−1)^n )/(3.2^n ))

$$\Rightarrow{ln}\left({a}_{{n}+\mathrm{1}} \right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({a}_{{n}} \right)−\frac{\mathrm{1}}{{z}}{ln}\left({a}_{{n}−\mathrm{1}} \right)=\mathrm{0} \\ $$$${ln}\left({a}_{{n}} \right).{w}_{{n}} \\ $$$$\Rightarrow{w}_{{n}+\mathrm{1}} −\frac{{w}_{{n}} }{\mathrm{2}}−\frac{{w}_{{n}−\mathrm{1}} }{\mathrm{2}}=\mathrm{0} \\ $$$${X}^{\mathrm{2}} −\frac{{X}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{0} \\ $$$${X}_{\mathrm{1}} =\mathrm{1},{X}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${w}_{{n}} ={a}+{b}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$${w}_{\mathrm{0}} =\mathrm{0}\Rightarrow{a}+{b}=\mathrm{0},\mathrm{2}{a}−{b}=\mathrm{2}{ln}\left(\mathrm{2}\right) \\ $$$${a}=\frac{\mathrm{2}}{\mathrm{3}}{ln}\left(\mathrm{2}\right),{b}=−\frac{\mathrm{2}}{\mathrm{3}}{ln}\left(\mathrm{2}\right) \\ $$$${w}_{{n}} ={ln}\left(\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)−\frac{\mathrm{2}}{\mathrm{3}}{ln}\left(\mathrm{2}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$={ln}\left(\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} .\mathrm{2}^{\frac{−\mathrm{2}}{\mathrm{3}}.\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }} \right) \\ $$$$={ln}\left(\mathrm{2}^{\frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{2}\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}.\mathrm{2}^{{n}} }} \right)={w}_{{n}} ,{a}_{{n}} ={e}^{{w}_{{n}} } =\mathrm{2}^{\frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{2}\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}.\mathrm{2}^{{n}} }} \\ $$$$ \\ $$

Commented by mr W last updated on 27/Jan/21

thank you sir! that′s what i also got: a_n =2^((2/3)(1−(((−1)^n )/2^n )))

$${thank}\:{you}\:{sir}! \\ $$$${that}'{s}\:{what}\:{i}\:{also}\:{got}: \\ $$$${a}_{{n}} =\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\right)} \\ $$

Commented by mindispower last updated on 27/Jan/21

always pleasur sir

$${always}\:{pleasur}\:{sir}\: \\ $$

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