a)\(\Rightarrow\frac{A}{2}=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{101}}\)
\(\Rightarrow A-\frac{A}{2}=\frac{1}{2}-\frac{1}{2^{101}}\)
\(\Rightarrow A=\frac{2^{100}-1}{2^{101}}\)
b)vì \(\frac{2^{100}}{2^{100}}=1\in N\Rightarrow\frac{2^{100}-1}{2^{100}}\ne1\notin N\left(đpcm\right)\)
A=\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\). Chứng minh rằng A\(\notin\)N
Rút gọn biểu thức sau:
a) S =\(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}\)
b) A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
AI LÀM NHANH MÀ ĐÚNG MIK TICK
chứng tỏ:
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{65}\notin N\)
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{40}}\notin N\)
\(C=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+.....+\frac{1}{61}\notin N\)
\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{9899}{9900}\)
CHO A=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\) 1/ RÚT GỌN A 2/ CHỨNG MINH A<1
1. Cho A=\(\frac{1}{1.101}\)+\(\frac{1}{2.102}\)+...+\(\frac{1}{10.110}\); B= \(\frac{1}{1.11}\)+\(\frac{1}{2.12}\)+...+\(\frac{1}{100.110}\)
Tính A : B
2. Cho A= 1+\(\frac{3}{2^3}\)+\(\frac{4}{2^4}\)+...+\(\frac{100}{2^{100}}\).Rút gọn A
Cho $A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\left(n\in Z;n\ge2\right)$A=142 +162 +182 +...+1(2n)2 (n∈Z;n≥2)
Chứng tỏ A$\notin$∉ N
a) A = 1+\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+..........+\frac{1}{100^2}\)
Chứng minh rằng A<2
b) B =\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+................+\frac{1}{2012^2}\)
Chứng minh rằng \(\frac{1}{2}-\frac{1}{2013}< B< 1\)
Rút gọn:
\(A=\frac{200+\frac{199}{2}+\frac{198}{3}+...+\frac{2}{199}+\frac{1}{200}}{\frac{100}{2}+\frac{100}{3}+...+\frac{100}{200}+\frac{100}{201}}\)
Bài 1
a rút gọn B=\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)
b Chứng minh A=\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}>\frac{5}{8}\)
