Question

Exercitiul 25,punctul a!Ajutor.............

Answer 1

$2+2*3+2*3^2+2*3^3+...+ 3^{2011}= x*81^{502}-1 \\\\ 2(1+3+3^2+...+3^{2011})=x*(3^4)^{502}-1 \\\\\\ \hbox{Pentru a rezolva o suma de puteri consecutive a unui numar se} \\ \hbox{foloseste formula sumei pentru o progresie geometrica}: \\\\ \boxed{S_n= b_1 * \frac{q^n-1}{q-1}} \\\\ \hbox{unde:} \\\\ S_n \to \hbox{suma primilor n termeni} \\ b_1 \to \hbox{primul termen al progresiei} \\ q \to \hbox{ratia progresiei} \\\\ b_1=1 \\ q=\frac{b_2}{b_1} \to \frac{3}{1} \to 3 \\\\ S_{2012}= 1* \frac{3^{2012}-1}{3-1}$

$S_{2012}= \frac{3^{2012}-1}{2} \\\\\\ \not2* \frac{3^{2012}-1}{\not2}= x * (3^{4})^{502}-1 \\\\ 3^{2012}-1= x* 3^{2008}-1 \\\\ 3^{2012}=x*3^{2008} \\\\ x= \frac{\not3^{2012}}{\not3^{2008}} \\\\\ \boxed{\boxed{x= 3^4 \to 81}}$

Answer 2

25. a).
Aplicam formula: 1+x+x²+ ... + xⁿ = (xⁿ⁺¹ -1)/(x-1) !!!
(1)... dau factor comun pe 2 ...
        =>  ...=2(1+3+3²+ ... +3²⁰¹¹)=x*(3⁴)⁵⁰² - 1
                   2* (3²⁰¹²-1)/(3-1)=x*3²⁰⁰⁸ - 1
                           3²⁰¹² - 1 = x*3²⁰⁰⁸ -1
                                     x = 3²⁰¹²:3²⁰⁰⁸=3⁴=81