Question

100 2a 2 12. Fie a = 20+1 +6+) 31+1 +99+1 unde ne N. Calculati 3 3
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Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it a=\dfrac{2^n}{3^n}:\dfrac{2\cdot2^n+6\cdot2^n\cdot3^n}{3\cdot3^n+9\cdot3^{2n}}=\dfrac{2^n}{3^n}\cdot\dfrac{3^n(3+9\cdot3^n)}{2^n(2+6\cdot3^n)}=\dfrac{3+9\cdot 3^n}{2+6\cdot3^n}=\\ \\ \\ =\dfrac{3(1+3^{n+1})}{2(1+3^{n+1})}=\dfrac{3}{2}$

$\it \Big(\dfrac{2a}{3}\Big)^{100}=\Big(\dfrac{2}{3}\cdot a\Big)^{100}=\Big(\dfrac{2}{3}\cdot \dfrac{3}{2}\Big)^{100}=1^{100}=1$