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Logaritmos
Enviado: Seg 10 Abr, 2017 20:08
por jomatlove
Se [tex3]x_{n}=\log \frac{3}{2}+\log \left(\frac{4}{3}\right)^{2}+\log \left(\frac{5}{4}\right)^{3}+...+\log \left(\frac{n+1}{n}\right)^{n-1}[/tex3]
,então o valor de:
[tex3]E=\frac{10^{-x_{n}}(n+1)^{n}}{n!}-n[/tex3]
a) [tex3]0[/tex3]
b) [tex3]1[/tex3]
c) [tex3]2[/tex3]
d) [tex3]\log 2[/tex3]
e) [tex3]\log n[/tex3]
Re: Logaritmos
Enviado: Ter 11 Abr, 2017 00:01
por Ittalo25
[tex3]10^{x_{n}} =10^{\log \frac{3}{2}+\log \left(\frac{4}{3}\right)^{2}+\log \left(\frac{5}{4}\right)^{3}+...+\log \left(\frac{n+1}{n}\right)^{n-1}} =[/tex3]
[tex3]10^{\log \frac{3}{2}}\cdot 10^{\log \left(\frac{4}{3}\right)^{2}} \cdot 10^{\log \left(\frac{5}{4}\right)^{3}} \cdot \cdot \cdot \cdot 10^{\log \left(\frac{n+1}{n}\right)^{n-1}}=[/tex3]
[tex3]\frac{3}{2} \cdot \left(\frac{4}{3}\right)^{2} \cdot \left(\frac{5}{4}\right)^{3} \cdot \cdot \cdot \cdot \left(\frac{n+1}{n}\right)^{n-1}= \frac{(n+1)^{n-1}}{n!}[/tex3]
Assim
[tex3]E=\frac{10^{-x_{n}}(n+1)^{n}}{n!}-n[/tex3]
[tex3]E=\frac{\frac{n!}{ (n+1)^{n-1}} \cdot (n+1)^{n}}{n!}-n[/tex3]
[tex3]E=n+1-n[/tex3]
[tex3]E=1[/tex3]