Các câu hỏi tương tự
Cho S=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
a) S =\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
b) \(\frac{7}{12}< S< \frac{5}{6}\)
cho A= 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/99.100
a, chứng tỏ : A= 1/51 + 1/52 + 1/53 + ... + 1/99.100
b, chứng tỏ 7/12< A< 5/6
A = \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\) . Chứng minh : \(\frac{7}{12}< A< \frac{5}{6}\)
Cho M= 1/2-3/4+5/6-7/8+...+197/198-199/200
N= 1/51+1/52+1/53+...+1/100
Tính N:M
cho S = 1/51 + 1/52 + 1/53 +....+ 1/99 + 1/100
chứng minh a, S < 1
b, S ko là số tự nhiên
CM: 1/51 +1/52+1/53 +................+1/100 < 5/6
tìm B=(1/2+1/6+1/12+...+1/9900)/(1/51+1/52+1/53+...+1/100)
M=1/2-3/4+5/6-7/8+...+197/198-199/200
N=1/51+1/52+1/53+...+1/100
Tính M : N
CMR :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{100}\)