Question

Exercitiul este in atasament , va rog!

Answer 1

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Answer 2

$\displaystyle a)~\sum\limits_{k=1}^n \ln \left(1- \frac{4}{(2k+1)^2} \right)= \sum\limits_{k=1}^n \ln \left( \left( 1+ \frac{2}{2k+1}\right) \left( 1- \frac{2}{2k+1}\right) \right)= \\ \\ = \sum\limits_{k=1}^n \left( \ln \left(1+ \frac{1}{2k+1} \right)+ \ln \left(1- \frac{2}{2k+1} \right) \right)= \\ \\ = \ \sum\limits_{k=1}^n \left( \ln \frac{2k+3}{2k+1}+ \ln \frac{2k-1}{2k+1}\right)= \\ \\ =  \sum\limits_{k=1}^n \ln \frac{2k+3}{2k+1}+  \sum\limits_{k=1}^n \ln \frac{2k-1}{2k+1}=$

$\displaystyle =  \left( \ln \frac{5}{3} + \ln \frac{7}{5} + \frac{9}{7}+...+ \ln \frac{2n+3}{2n+1}\right)+ \\ \\ +\left( \ln \frac{1}{3}+ \ln \frac{3}{5}+ \ln \frac{5}{7}+...+ \ln \frac{2n-1}{2n+1}\right)= \\ \\ = \ln \left( \frac{5}{3} \cdot \frac{7}{5} \cdot \frac{9}{7} \cdot ... \cdot \frac{2n+3}{2n+1} \right)+ \ln \left( \frac{1}{3} \cdot \frac{3}{5} \cdot \frac{5}{7} \cdot ... \cdot \frac{2n-1}{2n+1} \right)= \\ \\ = \ln \left( \frac{2n+3}{3}\right)+ \ln \left( \frac{1}{2n+1} \right)=$

$\displaystyle = \ln \left( \frac{2n+3}{3(2n+1)} \right) \to  \ln \frac{1}{3}=-\ln3.$

$\displaystyle b)~Bazat~pe~faptul~ca~ \left(1-a^{2^k}\right) \left(1+a^{2^k} \right)=1- \left( a^{2^k}\right)^2=1-a^{2 \cdot 2^k}= \\ \\ = 1-a^{2^{k+1}},~voi~inmulti~si~imparti~produsul~cu~ 1- \frac{1}{3^2}. \\ \\ a= \frac{1}{3} \\ \\ Produsul~va~fi: \frac{ \left(1-a^2 \right) \left(1+a^2 \right) \left(1+a^{2^2} \right) \left( 1+a^{2^3}\right)... \left(1+a^{2^n} \right)}{1-a^2}=$

$\displaystyle = \frac{ \left(1-a^{2^2} \right)\left(1+a^{2^2} \right) \left( 1+a^{2^3}\right)... \left(1+a^{2^n} \right)}{1-a^2}}= \\ \\ =\frac{ \left(1-a^{2^3} \right)\left(1+a^{2^3} \right) ... \left(1+a^{2^n} \right)}{1-a^2}}= \\ \\ =\frac{ \left(1-a^{2^4} \right) ... \left(1+a^{2^n} \right)}{1-a^2}}= \\ \\ ... \\ \\ = \frac{1-a^{2^{n+1}}}{1-a^2} \to \frac{1-0}{1-a^2}= \frac{9}{8}.$