CHỨNG MINH RẰNG :
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
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Chứng minh rằng \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
Ta có 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64
= ( 1/2 - 1/4 ) + ( 1/8 - 1/16 ) + ( 1/32 - 1/64)
= 1/4 + 1/16 + 1/64
= 16 + 4 + 1 /64
= 21/64 < 21/63 = 1/3
Vậy 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 < 1/3 ( đpcm ) Chúc bn hok tốt . k mik nha
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Chứng minh rằng:
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
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
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Chứng minh rằng \(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}<\frac{1}{3}\)
Chứng minh rằng :
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)\(\frac{1}{3}\)
27 tháng 1 2019 lúc 16:41
Chứng minh rằng:
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}<\frac{1}{3}\)
Giúp mình với!!!
Chứng minh rằng : \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}<\frac{1}{3}\)
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-...-\frac{1}{64}=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-..-\frac{1}{2^6}\) = 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 = \(1-\frac{1}{2^6}=\frac{2^6-1}{2^6}\)
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Đặt A = 1/2 − 1/4 + 1/8 − 1/16 + 1/32 − 1/64 A = 1/2 − 1/4 + 1/8 − 1/16 + 1/32 − 1/64
2A = 1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/322 A =1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/32
3A = 2A + A = 1 − 1/64 < 1 ⇒ A < 1/3 .
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Chứng Minh Rằng
a) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Chứng minh rằng:
a,\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b,\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
giúp minh với
Chứng minh rằng:
a) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
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)
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