Question

Se simplifică prin 6,pentru orice n€N ​

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Answer 1

$\frac{2^{n+3}+2^{n+2}+2^{n+1}+2^n}{3^{n+2}+2*3^{n+1}+3^n} = \frac{2^n*(2^3+2^2+2^1+1)}{3^n*(3^2*2*3^1+1)}=\frac{15*2^n}{16*3^n}= \\ \\ =\frac{2^{n-1}*2*3*5}{3^{n-1}*2*3*8} \rightarrow 2*3=6 \ se \ simplifica\\ \Rightarrow \frac{5*2^{n-1}}{8*3^{n-1}}$

Answer 2

$\it = \dfrac{2^{n-1}(16+8+4+2)}{3^{n-1}(27+18+3)}=\dfrac{\ 2^{n-1}\cdot30^{(6}}{3^{n-1}\cdot48}=\dfrac{2^{n-1}\cdot5}{3^{n-1}\cdot8}$