\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)
Bài 1 : Chứng minh
a) \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
b) \(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}...\frac{9999}{10000}< \frac{1}{100}\)
a) \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2014^2}< 2^{ }\)
b)\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
c)\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
a, A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}< 1\)
b, B=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}< 2\)
c, C=\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}< 6\)
d, D=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}< \frac{1}{100}\)
1)Chứng minh các phân số sau là các phân số tối giản:
a)\(A=\frac{12n+1}{30n+2}\)
b)\(B=\frac{14n+17}{21n+25}\)
2)Chứng minh rằng:
a)\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
b)\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
c)\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}>2\) \(C=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
\(B=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\) \(D=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}< \frac{1}{100}\)
Mọi người giúp mik nhé, mik đang ôn thi nên cần gấp!
chứng minh :
a,A=1+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{100^2}\)< 2
b,B=1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+...+\(\frac{1}{63}\)< 6
c,C=\(\frac{1}{2}\).\(\frac{3}{4}\).\(\frac{5}{6}\).\(\frac{7}{8}\)...\(\frac{9999}{10000}\)< \(\frac{1}{100}\)
Bài 1: Tìm x, biết:
\(\frac{1}{2.3}x+\frac{1}{3.4}x+\frac{1}{4.5}x+.....+\frac{1}{49.50}x=1\)
Bài 2: Chứng minh rằng:
\(a)A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{100^2}< 2\)
\(b)B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{63}< 6\)
\(c)C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.......\frac{9999}{10000}< \frac{1}{100}\)
Bài 3: Tính tổng:
\(S=\frac{1+2+2^2+2^3+.....+2^{2008}}{1-2^{2009}}\)
Chứng tỏ:
a) \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
b) \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{2^4}\right)....\left(1+\frac{1}{2^{50}}\right)< 3\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
d) \(\frac{1}{2}-\frac{1}{2^2}+.............+\frac{1}{2^{99}}-\frac{1}{2^{100}}< \frac{1}{3}\)
Chứng tỏ:
a) \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
b) \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{2^4}\right)...\left(1+\frac{2}{50}\right)< 3\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
d) \(\frac{1}{2}-\frac{1}{2^2}+.........+\frac{1}{2^{99}}-\frac{1}{2^{100}}< \frac{1}{3}\)
