Lời giải:
Ta có:
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow 2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow 2A-A=2-\frac{1}{2^{100}}\)
\(\Leftrightarrow A=2-\frac{1}{2^{100}}< 2\)
Vậy \(A< B\)
cho A=(\(\dfrac{1}{2^2}-1\))(\(\dfrac{1}{3^2}-1\))(\(\dfrac{1}{2^2}-1\))...........(\(\dfrac{1}{100^2}-1\)).SO sánh A với \(\dfrac{-1}{2}\)
Cho M = \(1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-\dfrac{1}{2^4}-....-\dfrac{1}{2^{10}}\) . So sánh M với \(\dfrac{1}{2^{11}}\)
So sánh:
a) A = \(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{49}}+\dfrac{1}{2^{50}}\) với 1
b) B = \(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\) với \(\dfrac{1}{2}\)
c) C = \(\dfrac{1}{4^1}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{999}}+\dfrac{1}{4^{1000}}\) với \(\dfrac{1}{3}\)
Cần gấp ạ ^^ Cảm ơn trước ^^
Tính :
1, A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+........+\dfrac{1}{100}\)
2, B = \(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.........+\dfrac{99}{100}\)
\(\text{Cho A=}\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^4}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}};\)
Hãy so sánh A với \(\dfrac{1}{3}\)
So Sánh : A = \(\dfrac{2009^{2009}+1}{2009^{2010}+1}\) và B = \(\dfrac{2009^{2010}-2}{2009^{2011}-2}\)
So Sánh : S = \(\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\) và \(\dfrac{1}{2}\)
Cho \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\) và \(B=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{200^2}\). Khi đó \(\dfrac{A}{B}=...\)
Cho B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{100^2}\). So sánh B với 3/4