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Question Number 199194 by hardmath last updated on 29/Oct/23

a_1 ,a_2 ,... a_1 =1 a_(2n) =n∙a_n a_2^(100) = ?

$$\mathrm{a}_{\mathrm{1}} ,\mathrm{a}_{\mathrm{2}} ,... \\ $$$$\mathrm{a}_{\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{a}_{\mathrm{2n}} =\mathrm{n}\centerdot\mathrm{a}_{\mathrm{n}} \\ $$$$\mathrm{a}_{\mathrm{2}^{\mathrm{100}} } \:=\:? \\ $$

Answered by mr W last updated on 29/Oct/23

a_2^n =a_(2×2^(n−1) ) =2^(n−1) a_2^(n−1) a_2^(n−1) =2^(n−2) a_2^(n−2) ... a_2^(2−1) =2^0 a_2^0 Π: a_2^n =2^((n−1)+(n−2)+...+2+1+0) a_1 =2^((n(n−1))/2) ⇒a_2^n =2^((n(n−1))/2) ⇒a_2^(100) =2^((100×99)/2) =2^(4950)

$${a}_{\mathrm{2}^{{n}} } ={a}_{\mathrm{2}×\mathrm{2}^{{n}−\mathrm{1}} } =\mathrm{2}^{{n}−\mathrm{1}} {a}_{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$${a}_{\mathrm{2}^{{n}−\mathrm{1}} } =\mathrm{2}^{{n}−\mathrm{2}} {a}_{\mathrm{2}^{{n}−\mathrm{2}} } \\ $$$$... \\ $$$${a}_{\mathrm{2}^{\mathrm{2}−\mathrm{1}} } =\mathrm{2}^{\mathrm{0}} {a}_{\mathrm{2}^{\mathrm{0}} } \\ $$$$\Pi: \\ $$$${a}_{\mathrm{2}^{{n}} } =\mathrm{2}^{\left({n}−\mathrm{1}\right)+\left({n}−\mathrm{2}\right)+...+\mathrm{2}+\mathrm{1}+\mathrm{0}} {a}_{\mathrm{1}} =\mathrm{2}^{\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}} \\ $$$$\Rightarrow{a}_{\mathrm{2}^{{n}} } =\mathrm{2}^{\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}} \\ $$$$\Rightarrow{a}_{\mathrm{2}^{\mathrm{100}} } =\mathrm{2}^{\frac{\mathrm{100}×\mathrm{99}}{\mathrm{2}}} =\mathrm{2}^{\mathrm{4950}} \\ $$

Commented by hardmath last updated on 29/Oct/23

perfect dear professor thank you

$$\mathrm{perfect}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$

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