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a-1-a-2-a-1-1-a-2n-n-a-n-a-2-100-




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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