\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{999^2}{999.1000}\)

\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.....\frac{999.999}{999.1000}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{999}{1000}\)

\(=\frac{1}{1000}\)

2y= 2/ 1.2.3 + 2/2.3.4 + 2/3.4.5 +.... +2/998.999.1000

2y=1/1.2 - 1/2.3 +1/2.3 - 1/3.4 + 1/3.4 -1/4.5 +....+ 1/998.999 - 1/ 999.1000

2y=1/2 - 1/ 999.1000

2y = 499500-1 /  999.1000

2y=499499 / 999.1000

y=499499 / 1998000

Ủng hộ mk nha

bai toan nay khó quá

Ta có : \(2Y=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+.....+\frac{2}{998.999.1000}\)

\(\Rightarrow2Y=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{998.999}-\frac{1}{999.1000}\)

\(\Rightarrow2Y=\frac{1}{1.2}-\frac{1}{999.1000}=\frac{1}{2}-\frac{1}{999000}=\frac{499499}{999000}\)

\(\Rightarrow Y=\frac{499499}{999000}\div2\)

\(C=\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+....+\frac{99.100-1}{100!}\)

\(\Rightarrow C=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(\Rightarrow C=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(\Rightarrow C=\left(2+\frac{3.4}{4!}+\frac{4.5}{5!}+....+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{10!}\right)\)

\(\Rightarrow C=\left(2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(\Rightarrow C=2-\frac{1}{99!}-\frac{1}{100!}< 2\Rightarrow C< 2\)

\(b,C=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+....+\frac{19}{9^2.10^2}\)

\(\Rightarrow C=\frac{3}{\left(1.2\right)\left(1.2\right)}+\frac{5}{\left(2.3\right)\left(2.3\right)}+...+\frac{19}{\left(9.10\right)\left(9.10\right)}\)

\(\Rightarrow C=\frac{3}{1.2}.\frac{1}{1.2}+\frac{5}{2.3}.\frac{1}{2.3}+....+\frac{19}{9.10}.\frac{1}{9.10}\)

\(\Rightarrow C=\left(1+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}+\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{3}\right)+....+\left(\frac{1}{9}+\frac{1}{10}\right)\left(\frac{1}{9}-\frac{1}{10}\right)\)

\(\Rightarrow C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{90}\)

\(\Rightarrow C=1-\frac{1}{90}< 1\Rightarrow C< 1\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{999\cdot1000}+1\)

\(=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+...+\frac{1000-999}{999\cdot1000}+1\)

\(=\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+...+\frac{1000}{999\cdot1000}-\frac{999}{999\cdot1000}+1\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(=1-\frac{1}{1000}+1\)

\(=\frac{999}{1000}+1\)

\(=\frac{1999}{1000}\)

1/1x2+1/2x3+1/3x4+...+1/99x100+1

1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100 +1

=1- 1/100 +1

=99/100 +1

=199/100

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/999.1000 + 1

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/999 - 1/1000 + 1

= 1 - 1/1000 + 1

= 2 - 1/1000

= 2000/1000 - 1/1000

= 1999/1000

Ủng hộ mk nha ♡_♡☆_☆

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)+1

\(=1-\frac{1}{1000}+1\)

=1-1+1/1000

= 1/1000

2, \(\frac{10}{1.2.3}+\frac{10}{2.3.4}+\frac{10}{3.4.5}+....+\frac{10}{100.101.102}\)

  \(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{102-100}{100.101.102}\)

  \(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\right)\)

  \(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{101.102}\right)\)

  \(=\frac{10}{2}.\frac{2575}{5151}\)

  \(=2,499514657\)

= 2,499514657 bạn nhé

1999/1000

= 1-1/2+1/2-1/3+1/3-1/4 + ... -1/999+1/999-1/1000 +1

= 1 - 1/1000 + 1 

= 1000/1001

Tk hộ mình nhé 

1 

Mà dấu . đó là gì

\(B=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{29.30}\right)\)

\(B=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{29}-\frac{1}{30}\right)\)

\(B=4.\left(1-\frac{1}{30}\right)\)

\(B=4.\frac{29}{30}\)

\(B=\frac{58}{15}\)

\(B=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{29}-\frac{1}{30}\right)\) 

\(=4\left(1-\frac{1}{30}\right)\) 

\(=4.\frac{29}{30}=\frac{58}{15}\) 

Vậy B= \(\frac{58}{15}\)

 \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)

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

\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)