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Ta có :
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
\(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}< \frac{1}{3}\)
\(\frac{16}{64}+\frac{4}{64}+\frac{1}{64}< \frac{1}{3}\)
\(\frac{16+4+1}{64}< \frac{1}{3}\)
\(\frac{21}{64}< \frac{1}{3}\)
=> 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 < 1/3
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-...-\frac{1}{64}=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-..-\frac{1}{2^6}\) = A
2A = 1 - 1/2 + 1/2^2 - ... - 1/2^5
2A + A = 1 - 1/2 + 1/2^2 - ... - 1/2^5 + 1/2 - 1/2^2 + 1/2^3 - 1/2^4 - .. - 1/2^6
3A = \(1-\frac{1}{2^6}=\frac{2^6-1}{2^6}\)
a)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
\(=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{32}-\frac{1}{64}\right)\)
\(=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}\)
\(=\frac{16+4+1}{64}\)
\(=\frac{21}{64}< \frac{1}{3}\)(đpcm)
Ta có: \(\frac{1}{2}-\frac{1}{4}=\frac{2}{4}-\frac{1}{4}=\frac{2-1}{4}=\frac{1}{4}\)
\(\frac{1}{8}-\frac{1}{16}=\frac{2}{16}-\frac{1}{16}=\frac{2-1}{16}=\frac{1}{16}\)
\(\frac{1}{32}-\frac{1}{64}=\frac{2}{64}-\frac{1}{64}=\frac{2-1}{64}=\frac{1}{64}\)
=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
=\(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}\)
=\(\frac{16}{64}+\frac{4}{64}+\frac{1}{64}=\frac{21}{64}\)
Ta có: \(\frac{21}{64}< \frac{21}{63}=\frac{1}{3}\)
=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
Đặt \(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)
\(A+2A=1-\frac{1}{64}\)
\(3A=1-\frac{1}{64}< 1\)
=>A<1/3
=>đpcm