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

$${\mathrm{a}}_1,{\mathrm{a}}_2,...$$$${\mathrm{a}}_1=1$$$${\mathrm{a}}_{2\mathrm{n}}=\mathrm{n}\cdot {\mathrm{a}}_{\mathrm{n}}$$$${\mathrm{a}}_{2^{100}}=?$$

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

$$a_{2^n}=a_{2\times 2^{n-1}}=2^{n-1}{a}_{2^{n-1}}$$$$a_{2^{n-1}}=2^{n-2}{a}_{2^{n-2}}$$$$...$$$$a_{2^{2-1}}=2^0{a}_{2^0}$$$$\mathrm{\Pi }:$$$$a_{2^n}=2^{(n-1)+(n-2)+...+2+1+0}{a}_1=2^{\frac{n(n-1)}{2}}$$$$\Rightarrow a_{2^n}=2^{\frac{n(n-1)}{2}}$$$$\Rightarrow a_{2^{100}}=2^{\frac{100\times 99}{2}}=2^{4950}$$

Commented by hardmath last updated on 29/Oct/23

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

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