ta có A = \(\frac{1.4}{2.3}+\frac{2.5}{3.4}+....+\frac{98.101}{99.100}=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{4950}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{6}+....+\frac{1}{4950}\right)\)(có 98 chữ số 1)
\(=98-\left(\frac{1}{3}+\frac{1}{6}+....+\frac{1}{4950}\right)\)=> A < 98
đi rùi giải tiếp
Cho \(N=\frac{1.4}{2.3}+\frac{2.5}{3.4}+\frac{3.6}{4.5}+...+\frac{98.101}{99.100}\)
Chứng minh : 97 < N < 98
Cho \(N=\frac{1.4}{2.3}+\frac{2.5}{3.4}+\frac{3.6}{4.5}+...+\frac{98.101}{99.100}\)
Chứng minh : 97 < N < 98
TÍNH N=\(\frac{1.4}{2.3}+\frac{2.5}{3.4}+...+\frac{98.101}{99.100}\)
Cho N = 1.4/2.3 + 2.5/3.4 + 3.6/4.5 + ... + 98.101/99.100 . Chứng minh 97 < N < 98
a)CHO A = \(\frac{2}{3^2}\)+ \(\frac{2}{5^2}\) + \(\frac{2}{7^2}\)+.........+ \(\frac{2}{2007^2}\)
CHỨNG MINH RẰNG: A < \(\frac{1003}{2008}\)
b)CHO N = \(\frac{1.4}{2.3}\)+ \(\frac{2.5}{3.4}\)+ \(\frac{3.6}{4.5}\)+.........+ \(\frac{98.101}{99.100}\)
CHỨNG MINH RẰNG: 97 < N < 98
~ giúp mk với~
biết a = 1.2+2.3+3.4+.....+99.100=333300 thì B = 1.4+2.5+3.6+....+99.102=......
biết A = 1.2+2.3+3.4+.........+99.100=333300 thì B = 1.4+2.5+3.6+......+99.102 = .....
Biết A=1.2+2.3+3.4......+99.100=333300 thì B=1.4+2.5+3.6..+99.102= ?
Biết A=1.2+2.3+3.4+...+99.100=333300 thì B=1.4+2.5+3.6+...+ 99.102=?
