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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
Lời giải:
$M=\frac{1.4}{2.3}+\frac{2.5}{3.4}+\frac{3.6}{4.5}+...+\frac{98.101}{99.100}$
$=1-\frac{2}{2.3}+1-\frac{2}{3.4}+1-\frac{2}{4.5}+...+1-\frac{2}{99.100}$
$=(1+1+....+1)-2(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100})$
$=98-2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100})$
$=98-2(\frac{1}{2}-\frac{1}{100})$
$=97+\frac{1}{50}=97,02$
E=\(\frac{1.2.3.....97.98}{2.3.4.....98.99}\)+\(\frac{4.5.6....100.101}{3.4.5...99.100}\)
E=\(\frac{1}{99}\)+\(\frac{101}{3}\)
E=\(\frac{304}{99}\)
\(\text{Ta có: }\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{99.100}\)
\(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)
\(=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=5.\left(1-\frac{1}{100}\right)\)
\(=5.\frac{99}{100}\)
\(=\frac{99}{20}\)
5/1.2 + 5/2.3 + 5/3.4 + ... + 5/99.100
= 5 . ( 1/1.2 + 1/2.3 + 1/3.4 +... + 1/99.100 )
= 5 . ( 1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100 )
= 5 . ( 1 - 1/100 )
= 5 . 99/100
= 99/20