Question

Scrieti ca putere a lui a, unde n= numarul radicalilor.
n∈ N, n≥2

Answer 1

$c)\quad \sqrt{a\cdot \sqrt[3]{a^2\cdot \sqrt[4]{a^3\cdot ...\cdot \sqrt[n-1]{a^{n-2\cdot }\sqrt[n]{a^{n-1}}}}}}=$

$= \left(...\left(\left(\left(\left(a^{n-1}\right)^{\frac{1}{n}}\cdot a^{n-2}\right)^{\frac{1}{n-1}}\cdot a^{n-3}\right)^{\frac{1}{n-2}}\cdot a^{n-4}\right)^{\frac{1}{n-3}}\cdot ...\cdot a\right)^{\frac{1}{2}} =\\ \\ \\= a^{\left(...\left(\left(\left(\frac{n-1}{n}+n-2\right)\frac{1}{n-1}+n-3\right)\frac{1}{n-2}+n-4\right)\frac{1}{n-3}\cdot ...\cdot 1 \right)\frac{1}{2}} \overset{*}{=}$

$\\\text{Acum calculez puterile separat, la rand, de la stanga la dreapta.}$

$\Big(\dfrac{n-1}{n}+n-2\Big)\dfrac{1}{n-1} = \Big(\dfrac{n-1}{n}+n-1-1\Big)\dfrac{1}{n-1}=\\ \\ = (n-1)\cdot \Big(\dfrac{1}{n}+1-\dfrac{1}{n-1}\Big)\cdot \dfrac{1}{n-1} = \\ \\ = \dfrac{1}{n}-\dfrac{1}{n-1}+1 = \dfrac{n-1-n}{n(n-1)}+1 = -\dfrac{1}{n(n-1)}+1\\\\\\\text{Continuam:}\\\\\Big(-\dfrac{1}{n(n-1)}+1+n-3\Big)\dfrac{1}{n-2} = \\ \\ =\Big(-\dfrac{1}{n(n-1)}+n-2\Big)\dfrac{1}{n-2} = -\dfrac{1}{n(n-1)(n-2)}+1\\ \\\\\text{Continuand, vom obtine si:} \\ \\ -\dfrac{1}{n(n-1)(n-2)(n-3)}+1$

$\\\text{Analog:}$

$\overset{*}{=}\,a^{\displaystyle -\frac{1}{n\cdot (n-1)\cdot (n-2)\cdot (n-3)\cdot ...\cdot 2}+1} = \\ \\ =a^{\displaystyle -\dfrac{1}{n!}+1} = \boxed{a^{\displaystyle 1 - \dfrac{1}{n!}}}$

Answer 2

Am atasat o rezolvare.