Chứng tỏ rằng:
a, \dfrac{1}{2} – \dfrac{1}{4} + \dfrac{1}{8} – \dfrac{1}{16} + \dfrac{1}{32} – \dfrac{1}{64} < \dfrac{1}{3}
b, \dfrac{1}{41} + \dfrac{1}{42} + \dfrac{1}{43} + ... + \dfrac{1}{79} + \dfrac{1}{80} > \dfrac{7}{12}
Chứng tỏ rằng:
a, \dfrac{1}{2} – \dfrac{1}{4} + \dfrac{1}{8} – \dfrac{1}{16} + \dfrac{1}{32} – \dfrac{1}{64} < \dfrac{1}{3}
b, \dfrac{1}{41} + \dfrac{1}{42} + \dfrac{1}{43} + ... + \dfrac{1}{79} + \dfrac{1}{80} > \dfrac{7}{12}
b) Đặt A= $\frac{1}{2}$- $\frac{1}{4}$+ $\frac{1}{8}$- $\frac{1}{16}$+ $\frac{1}{32}$- $\frac{1}{64}$
2A= 1- $\frac{1}{2}$+ $\frac{1}{4}$- $\frac{1}{8}$+ $\frac{1}{16}$- $\frac{1}{32}$
2A- A= 1- $\frac{1}{64}$
A= $\frac{63}{64}$
Vì $\frac{63}{64}$< $\frac{1}{3}$
nên $\frac{1}{2}$- $\frac{1}{4}$+ $\frac{1}{8}$- $\frac{1}{16}$+ $\frac{1}{32}$- $\frac{1}{64}$< $\frac{1}{3}$
Vậy $\frac{1}{2}$- $\frac{1}{4}$+ $\frac{1}{8}$- $\frac{1}{16}$+ $\frac{1}{32}$- $\frac{1}{64}$< $\frac{1}{3}$
b)
$\frac{7}{12}$= $\frac{4}{12}$+ $\frac{3}{12}$= $\frac{20}{60}$+ $\frac{20}{80}$
⇒
= ( $\frac{1}{41}$+$\frac{1}{42}$+ $\frac{1}{43}$+…+ $\frac{1}{60}$)+ ($\frac{1}{61}$+ $\frac{1}{62}$+…+ $\frac{1}{79}$+ + $\frac{1}{80}$
Vì $\frac{1}{41}$+$\frac{1}{42}$+ $\frac{1}{43}$> …> $\frac{1}{59}$+$\frac{1}{60}$
⇒ $\frac{1}{41}$+$\frac{1}{42}$+ $\frac{1}{43}$+…+$\frac{1}{60}$)> $\frac{1}{60}$+…+ $\frac{1}{60}$= $\frac{20}{60}$
và $\frac{1}{61}$>$\frac{1}{62}$…> $\frac{1}{79}$> $\frac{1}{80}$
⇒ ($\frac{1}{61}$+ $\frac{1}{62}$+…+ $\frac{1}{79}$+ + $\frac{1}{80}$)> $\frac{1}{80}$>… $\frac{1}{80}$= $\frac{20}{80}$
Vậy $\frac{1}{41}$+$\frac{1}{42}$+ $\frac{1}{43}$+…+$\frac{1}{79}$+ + $\frac{1}{80}$> $\frac{20}{60}$+ $\frac{20}{80}$= $\frac{7}{12}$
⇒ $\frac{1}{41}$+$\frac{1}{42}$+ $\frac{1}{43}$+…+ $\frac{1}{79}$+ $\frac{1}{80}$> 7/12
CHO MIK CTLHN NHA