Ta có 1.4/2.3=(2-1)(3+1)/2.3=1-1/2+1/3-1/2.3

2.5/3.4=(3-1)(4+1)/3.4=1-1/3+1/4-1/3.4

...

Suy ra N=(1-1/2+1/3-1/2.3)+(1-1/3+1/4-1/3.4)+....+(1-1/99+1/100-1/99.100)

N=98+1/100−1/2−1/2.3−1/3.4−....−1/99.100

Xét P=1/2.3+1/3.4+....+1/99.100

P= 1/2−1/3+1/3−1/4+.....+1/99−1100 

P=1/2−1/100

Vậy N=98-1+1/50

N=97+1/50

Vậy 97<N<98(ĐPCM)

giải chưa nhở

mk cx đg cần giải bài này

N = 1 - 2/2.3 + 1 - 2/3.4 +.....+ 1 - 2/99.100

   = 98 - 2.(1/2.3 + 1/3.4 + ...... + 1/99.100)

   = 98 - 2.(1/2-1/3+1/3-1/4+....+1/99-1/100)

   = 98 - 2.(1/2-1/100)

   = 98 - 2.49/100 = 98-49/50 < 98

Mà 49/50 < 1

=> N > 98-1 = 97

=> 97 < N < 98

Tk mk nha

Ta có 1.4/2.3=(2-1)(3+1)/2.3=1-1/2+1/3-1/2.3

2.5/3.4=(3-1)(4+1)/3.4=1-1/3+1/4-1/3.4

...

Suy ra N=(1-1/2+1/3-1/2.3)+(1-1/3+1/4-1/3.4)+....+(1-1/99+1/100-1/99.100)

N=\(98+\dfrac{1}{100}-\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-....-\dfrac{1}{99.100}\)

Xét P=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)

P=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)

P=\(\dfrac{1}{2}-\dfrac{1}{100}\)

Vậy N=98-1+\(\dfrac{1}{50}\)

N=\(97+\dfrac{1}{50}\)

Vậy 97<N<98(ĐPCM)

Ta có công thức tổng quát của số hạng trong tổng trên có dạng:

\(x_n=\frac{n\left(n+3\right)}{\left(n+1\right)\left(n+2\right)}=\frac{n^2+3n+2-2}{n^2+3n+2}\)

\(=1-\frac{2}{n^2+3n+2}=1-\frac{2}{\left(n+1\right)\left(n+2\right)}\)

\(\Rightarrow\frac{1.4}{2.3}=1-\frac{2}{2.3}\)

\(\frac{2.5}{3.4}=1-\frac{2}{3.4}\)

\(\frac{3.6}{4.5}=1-\frac{2}{4.5}\)

....

\(\frac{98.101}{99.100}=1-\frac{2}{99.100}\)

\(\Rightarrow N=98-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

\(=98-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=98-2\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=98-1+\frac{1}{50}=97+\frac{1}{50}\)

Vậy 97 < N < 98

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N=\(\frac{1.4}{2.3}+\frac{2.5}{3.4}+\frac{3.6}{4.5}+....+\)\(\frac{98.101}{99.100}\)

N=\(\frac{1.2.3...98}{2.3.4...99}\)\(+\)\(\frac{4.5.6....101}{3.4.5....100}\)

N=\(\frac{1}{99}+\frac{101}{3}\)

N=\(\frac{3334}{99}\)