\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018+2019}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
=\(1-\frac{1}{2019}< 1\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(A=\frac{1}{1}-\frac{1}{2019}< 1\)
Vậy \(A< 1\)
Cho:
A=1/1.2+1/2.3+1/3.4+....+1/2016.2017.Chứng tỏ A<1
chứng tỏ rằng: 1/1.2+1/2.3+1/3.4+...+1/49.50<1
Chứng tỏ rằng: 1/1.2+1/2.3+1/3.4+....+1/99.100<1
chứng tỏ rằng:1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 < 1
Chứng tỏ rằng: 1/1.2+1/2.3+1/3.4+...+1/1999.2000<1
chứng tỏ
1/1.2+1/2.3+1/3.4+.....+1/99.100<1
Chứng tỏ rằng: 1/1 - 1/2 =1/1.2 ; 1/2 -1/3 = 1/2.3 ; 1/3 -1/4 =1/3.4
Tính S = 1/1.2 + 1/2.3 + 1/3.4+... 1/49.50
Chứng tỏ rằng :
a) 1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/99.100 < 1
b) 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2
Chứng tỏ :
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50 = 1/26 + 1/27 + .. + 1/ 50