Question

bună ziua! mă poate ajuta cineva la acest ex?​

Answer 1

a)

$\it x=14^{n+1}(26+42-35)=33\cdot14^{n+1}=11\cdot3\cdot15^{n+1}\in M_{11} \Rightarrow x\ \vdots\ 11$

b)

$\it 33\cdot14^{n+1} < 33\cdot196^3\Big|_{:33} \Rightarrow 14^{n+1} < 14^6 \Rightarrow n+1 < 6 \Rightarrow \\ \\ \Rightarrow n < 5 \Rightarrow n\in\{0,\ 1,\ 2,\ 3,\ 4\}$

Answer 2

Răspuns:

a) b) n ∈ {1, 2, 3, 4}

Explicație pas cu pas:

a)

$x = 13 \cdot {2}^{n + 2} \cdot {7}^{n + 1} + 3 \cdot {14}^{n + 2} - 5 \cdot {2}^{n + 1} \cdot {7}^{n + 2} = 13 \cdot 2 \cdot {2}^{n + 1} \cdot {7}^{n + 1} + 3 \cdot 14 \cdot {14}^{n + 1} - 5 \cdot {2}^{n + 1} \cdot 7 \cdot {7}^{n + 1} = 26 \cdot {(2 \cdot 7)}^{n + 1} + 42 \cdot {14}^{n + 1} - 35 \cdot {(2 \cdot 7)}^{n + 1} = 26 \cdot {14}^{n + 1} + 42 \cdot {14}^{n + 1} - 35 \cdot {14}^{n + 1} = {14}^{n + 1} \cdot (26 + 42 - 35) = {14}^{n + 1} \cdot 33 = {14}^{n + 1} \cdot 3 \cdot 11 \: \: \red{ \bf \vdots \ 11}$

b)

$x < 33 \cdot {196}^{3}$

${14}^{n + 1} \cdot 33 < 33 \cdot {196}^{3} \\ {14}^{n + 1} < {( {14}^{2} )}^{3} \iff {14}^{n + 1} < {14}^{6} \\ n + 1 < 6 \implies n < 5$

=> n ∈ {1, 2, 3, 4}