Question

13) Fie a=1+3+3^2+.....+3^10+3^11
a) Arătați ca a e nr par
b) Arătați ca nr a e Divizibil cu 10

Answer 1

13.

$a = 1+3+3^2+3^3+\cdots + 3^10+3^11\\ \\ \textrm{Progresie geometrica}\\ \\ b_1 = 1\\ \\ q = 3\\ \\ n = 12 \\ \\ S_n = S_{12} = \frac{1(q^{12} - 1)}{q - 1} = \frac{3^{12} - 1}{3 - 1} = \frac{3^{12}- 1}{2} \\ \\ a \in M_2 \iff \frac{3^{12} - 1}{2} = 2k \iff 3^{12} - 1 = 4k \iff \frac{3^{12}-1}{4} = k \iff k = \frac{(3^6+1)(3^6-1)}{4} \\ \\ k = \frac{(3^6+1)(3^3+1)(3^3-1)}{4} = \frac{(3^6+1)\cdot 28\cdot 26}{4} = (3^6+1)\cdot 7 \cdot 26\in \mathbb{Z}\implies a \in M_2$

$10 \mid a \\ a = 10k \\ \\ \frac{3^{12}- 1}{2} = 10k \iff k = \frac{3^{12} - 1}{20}\\ \\ k = \frac{(3^6 + 1)(3^3+1)(3^3-1)}{20} = \frac{(3^6+1)\cdot 28\cdot 24}{20} = \frac{(3^6 + 1)\cdot 14\cdot 12}{5}\\ \\ 3^6 = {(3^3)}^2 = 27^2 = (25+2)^2 = 625 + 100 + 4 = 729\\ \\ 3^6 + 1 = 730 \in M_5 \implies \frac{730 \cdot 14\cdot 12}{5} \in \mathbb{Z} \implies a \in M_{10}$

Answer 2

$(a+b)^n = M_{a}+b^n$

a)

$a = 1+3+3^2+...+3^{10}+3^{11} \\ a = 1+(2+1^1)+(2+1)^2+...+(2+1)^{10}+(2+1)^{11} \\ a = 1+(2+1^1)+(M_2+1^2)+...+(M_{2}+1^{10})+(M_2+1^{11}) \\ a =1+1^2+1^2+...+1^{10}+1^{11}+M_2\\ a = 1\cdot 12+M_2 \\ a = M_2+M_2 \\ a=M_2\\ \\\Rightarrow a \text{ par}$

b)

$a = 1+3+3^2+3^3+...+3^{10}+3^{11} \\3a = \quad\,3+3^2+3^3+....+3^{10}+3^{11}+3^{12} \\ \noindent\rule{5.9cm}{0.7pt} \\ 3a-a =3^{12}-1 \\ 2a = 3^{12}-1 \\ \\a = \dfrac{3^{12}-1}{2} = \dfrac{27\cdot 3^{9}-1}{2} = \dfrac{(20+7)3^{9}-1}{2} = \dfrac{M_{20}+7\cdot 3^{9}-1}{2} = \\ \\ =M_{10}+\dfrac{7\cdot 3\cdot 3^8-1}{2} = M_{10}+\dfrac{(20+1)\cdot 3^8 - 1}{2} = M_{10}+\dfrac{3^8-1}{2} =\\\\ \text{Analog (observam ca puterea scade din 4 in 4):} \\\\ =M_{10}+\dfrac{3^0 - 1}{2} = M_{10}$