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catalina17
Pe: 2012-07-10T14:51:29+03:00 In: In:

patrat perfect

Fie A=(1/1*3 + 1/3*5 + 1/5*7 + … + 1/2007*2009) : 2008/2009 * 2^(4n+7)

a. Aratati ca A este patrat perfect pt oricare n apartine N
b. Determinati n apartine N*, astfel incat A=64^(n-1)

Am facut asa:

1 – 1/3 + 1/3 – 1/5 + 1/5 -1/7+ … + 1/2007 – 1/2009 = 1 – 1/2009 = 2008/2009 ❗

2008/2009 : 2008/2009 * 2^(4n+7)

Mai departe nu mai stiu ce sa fac, va rog daca puteti sa ma ajutati!
Multumesc!

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1 raspuns

  1. dan
    expert (VI)

    ❗
    \rm{\fbox{\frac{1}{k(k+2)}=(\frac{1}{k}-\frac{1}{k+2}) \cdot \frac{1}{2}=\frac{k+2-k}{k(k+2)} \cdot \frac{1}{2} } \\ \Rightarrow \{\frac{1}{1*3} = \frac{1}{2} \cdot (\frac{1}{1}-\frac{1}{3}) \\ \frac{1}{3*5} = \frac{1}{2} \cdot (\frac{1}{3}-\frac{1}{5}) \\ ...............................\\ \underline{\frac{1}{2007*2009}=\frac{1}{2} \cdot (\frac{1}{2007}-\frac{1}{2009}) } \\ ( ... ) = \frac{1}{2} \cdot (\frac{1}{1}-\frac{1}{2009})=\frac{1}{2} \cdot \frac{2008}{2009} \surd \bl

    \rm{a). \Rightarrow A=\frac{1}{2}\cdot \frac{2008}{2009} : \frac{2008}{2009}\cdot2^{4n+7}=\frac{1}{2}\cdot\frac{2008}{2009}\cdot\frac{2009}{2008}\cdot\frac{2^{4n+8}}{2}=\frac{(2^{2n+4})^2}{2^2}=\(\frac{(2^{2n+4})}{2}\)^2=p.p. \forall n\in{N}\surd\surd\\ b). daca A= \underbrace{ 64^{n-1} = \(\frac{(2^{2n+4})}{2}\)^2 }_{n=? \in{ N* }} \Leftrightarrow (2^6)^{n-1} \cdot 2^2= 2^{4n+8} \Leftrightarrow 2^{6n-6+2}=2^{4n+8}\\ deci 6n-4=4n+8 de unde se obtine n= 6 \in \mathbb{N^*}\bl

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