Câu A
Ta có (1/2)A = 1/22 + 1/23 + ... + 1/2100 + 1/2101
=> (1/2)A - A = - (1/2)A = (1/22 + 1/23 + ... + 1/2100 + 1/2101) - (1/2 + 1/22 + ... + 1/2100 )
= 1/2101 - 1/2
=> A = 1 - 1/2100
Câu B
Ta có 1/(1x2) = 1/1 - 1/2
1/(2.3) = 1/2 - 1/3
.................................
1/(99.100) = 1/99 - 1/100
=> B = 1/1 - 1/2 + 1/2 - 1/3 +.... +1/99 - 1/100
= 1 - 1/100
=99/100
\(\frac{1\times2}{2\times3}+\frac{2\times3}{3\times4}+\frac{3\times4}{4\times5}+...+\frac{98\times99}{99\times100}\)
Tính biểu thức A
\(A=\frac{5}{1\times2}+\frac{5}{2\times3}+\frac{5}{3\times4}+...+\frac{5}{98\times99}+\frac{5}{99\times100}\)
Cho A=\(\frac{1\times2-1}{2!}+\frac{2\times3-1}{3!}+\frac{3\times4-1}{4!}+.....+\frac{99\times100-1}{100!}<2\)
Tính:
\(A=\frac{1^2}{1\times2}\times\frac{2^2}{2\times3}\times\frac{3^2}{3\times4}\times...\times\frac{99^2}{99\times100}\times\frac{100^2}{100\times101}\)
\(y=\frac{1\times100+2\times99+3\times98...+99\times2+100\times1}{1\times2+2\times3+3\times4+...+99\times100+100\times101}=?\)
\(B=\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{\frac{6}{5}+\frac{6}{7}-\frac{2}{3}+\frac{6}{11}}+\frac{\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right)\times2021}{\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}}\) Giá trị của B là:(cho mình cách giải chi tiết nhé mình sẽ tick)
Tính :\(A=\frac{5}{1\times2}+\frac{5}{2\times3}+\frac{5}{3\times4}+....+\frac{5}{99\times100}\)Ai nhanh mik tick nha^_-
Cho A =\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{99\times100}\)
Chứng minh rằng \(\frac{5}{6}\)< A < \(\frac{7}{12}\)
Tính giá trị biểu thức A = \(\frac{1}{1\times2}+\frac{1}{2\times3}+...\frac{1}{99\times100}\)
