Question

.........ex 30 va rog!!! dau coroana si 50 pct

Answer 1

a)

$\it a_n=a_1q^{n-1}\\ \\ a_1\cdot a_2\cdot a_3\cdot\ ...\ \cdot a_n=a_1\cdot a_1\cdot q\cdot a_1\cdot q^2\cdot\ ...\ \cdot a_1\cdot q^{n-1}=a_1^n\cdot q^{\dfrac{(n-1)n}{2}}=\\ \\ = a_1^{\dfrac{2n}{2}}\cdot q^{\dfrac{(n-1)n}{2}} =\sqrt{a_1^{2n}\cdot q^{(n-1)n}}=\sqrt{(a_1^2\cdot q^{n-1})^n} =\sqrt{(a_1\cdot a_1\cdot q^{n-1})^n} =\\ \\ =\sqrt{(a_1\cdot a_n)^n}$

b)  

$\it \dfrac{\sqrt{a_n}}{\sqrt{a_n}-\sqrt{a_{n-1}}} =\dfrac{\sqrt{a_n}(\sqrt{a_n}+\sqrt{a_{n-1}})}{a_n-a_{n-1}}=\dfrac{a_n+\sqrt{a_n\cdot a_{n-1}}}{a_n-a_{n-1}}=\\ \\ \\ =\dfrac{a_{n -1}\cdot q+\sqrt{a_{n -1}\cdot q\cdot a_{n-1}}}{a_{n-1}\cdot q-a_{n-1}}=\dfrac{a_{n-1}(q+\sqrt q)}{a_{n-1}(q-1)}=\dfrac{q+\sqrt q}{q-1}=\\ \\ \\ = \dfrac{\sqrt q(\sqrt q+1)}{(\sqrt q)^2-1}=\dfrac{\sqrt q(\sqrt q+1)}{(\sqrt q-1)(\sqrt q+1)}=\dfrac{\sqrt q}{\sqrt q-1}\ \ \ \ (*)$

Suma din enunț are (n-1) termeni, fiecare termen reducându-se

la forma (*).

Deci, suma va fi egală cu :

$\it \dfrac{(n-1)\sqrt q}{\sqrt q-1}$