Đặt \(B=1+\dfrac{1}{2}+...+\dfrac{1}{1024}\) và \(A=-1-\dfrac{1}{2}-\dfrac{1}{4}-...-\dfrac{1}{1024}\)
=>A=-B
\(B=1+\dfrac{1}{2}+...+\dfrac{1}{1024}\)
=>\(\dfrac{1}{2}B=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^{11}}\)
=>\(-\dfrac{1}{2}B=\dfrac{1}{2^{11}}-1\)
=>\(\dfrac{1}{2}B=1-\dfrac{1}{2^{11}}=\dfrac{2^{11}-1}{2^{11}}\)
=>\(B=\dfrac{2^{11}-1}{2^{10}}\)
=>\(A=\dfrac{1-2^{11}}{2^{10}}\)
-1-\(\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{1024}\)
\(\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{1024}\)
tính giá trị của biểu thức : B\(=\dfrac{\left(\dfrac{1}{2}\right)^{10}.5-\left(\dfrac{1}{4}\right)^5.3}{\dfrac{1}{1024}.\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\)
a) 2\(\dfrac{2}{7}\)x - \(\dfrac{5}{8}\)=\(\dfrac{17}{24}\)
b) -1 - \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) . . . \(\dfrac{1}{1024}\)
c) \(1\dfrac{1}{24}\). \(3\dfrac{3}{8}\) : \(3\dfrac{3}{4}\)
Tính tổng đại số
\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)
\(tính:\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{2}\)
8) \(A=\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
9) \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2014}}+\dfrac{1}{3^{2015}}\)
10) \(P=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2005}}{\dfrac{2004}{1}+\dfrac{2003}{2}+\dfrac{2002}{3}+...+\dfrac{1}{2004}}\)
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
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 ^^
