Question

CM rằng tổng P=\(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+.......+\frac{1}{3^{2006}}-\frac{1}{3^{2008}}< 0,1\)

Giúp mk vs nak!!!!!!!!!!!!!!!

Thanks nhìu nok!!!!!!!!!

Answer 1

9P = 1 - \(\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+.....................+\frac{1}{3^{2004}}-\frac{1}{3^{2006}}\)

9P + P = \(\left(1-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+.....................+\frac{1}{3^{2004}}-\frac{1}{3^{2006}}\right)\)+ \(\left(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+........................+\frac{1}{3^{2006}}-\frac{1}{3^{2008}}\right)\)

10P = 1 - \(\frac{1}{3^{2008}}\)

Suy ra : P = \(\frac{1}{10}-\frac{1}{3^{2008}.10}\)

Vì \(\frac{1}{3^{2008}.10}>0\) nên \(\frac{1}{10}-\frac{1}{3^{2008}.10}< \frac{1}{10}\) hay P < 0,1 ( ĐPCM)