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Question Number 167955 by peter frank last updated on 30/Mar/22

Answered by JDamian last updated on 30/Mar/22

6^((1/2^1 )+(1/2^2 )+(1/2^3 )+∙∙∙) =6^1 =6

$$\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\centerdot\centerdot\centerdot} =\mathrm{6}^{\mathrm{1}} =\mathrm{6} \\ $$

Answered by alephzero last updated on 30/Mar/22

6^((1/2)+(1/4)+(1/8)+...)  = exp_6 (Σ_(n=1) ^∞ (1/2^n ))  S = (a_1 /(1−q))     a_1  = (1/2) ∧ q = (1/2)  ⇒ S = ((((1/2)))/((1−(1/2)))) = 1  ⇒ 6^(1/2) 6^(1/4) 6^(1/8) ... = 6^1  = 6          ■ QED

$$\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+...} \:=\:\mathrm{exp}_{\mathrm{6}} \left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right) \\ $$$${S}\:=\:\frac{{a}_{\mathrm{1}} }{\mathrm{1}−{q}}\:\:\:\:\:{a}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\wedge\:{q}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:{S}\:=\:\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{6}^{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{6}^{\frac{\mathrm{1}}{\mathrm{8}}} ...\:=\:\mathrm{6}^{\mathrm{1}} \:=\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\blacksquare\:{QED} \\ $$

Commented by peter frank last updated on 30/Mar/22

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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