tui chịu

Ta có: \(\frac{3}{5\cdot2!}+\frac{3}{5\cdot3!}+\frac{3}{5\cdot4!}+.....+\frac{3}{5\cdot99!}+\frac{3}{5\cdot100!}\)

= \(\frac{3}{5}\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)

\(< \frac{3}{5}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

= \(\frac{3}{5}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{3}{5}\left(1-\frac{1}{100}\right)< \frac{3}{5}=0,6\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)

k=2

chuan 100%

k=2 do

Đặt A =  \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+.....+\frac{3}{99.100}\)

\(\frac{1}{3}A\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)

\(\frac{1}{3}A\)\(=1-\frac{1}{100}\)

=> \(\frac{1}{3}A=\frac{99}{100}\)

=> A = \(\frac{99}{100}.3=\frac{297}{100}\)

     \(\frac{3}{1.2}+\frac{3}{2.3}+..................+\frac{3}{99.100}\)

\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+..................+\frac{1}{99.100}\right)\)

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

\(=3.\left(1-\frac{1}{100}\right)\)

\(=3.\frac{99}{100}\)

\(=\frac{297}{100}\)

\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)

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

\(=\frac{3}{1}-\frac{3}{100}\)

\(=\frac{297}{100}\)

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

\(\Rightarrow\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(=\frac{1}{1.2}-\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{1.2.3}+...+\frac{1}{98.99.100}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Rightarrow k=2\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)

\(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)

Ta có : \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75}\), \(\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)nên :

\(A>\frac{1}{75}.25+\frac{1}{100}.25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

\(A< \frac{1}{51}.25+\frac{1}{76}.25< \frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

Vậy \(\frac{7}{12}< A< \frac{5}{6}\)

+A=1/(1.2)+1/(3.4)+...+1/(99.100) 
=1/1-1/2+1/3-1/4+....+1/99-1/100 
=1/2+1/3-1/4+1/5-1/6+1/7+...-1/98+1/99... 
=(1/2+1/3)+(1/5-1/4)+(1/7-1/6)+..+(1/9... 
=5/6-(1/4.5+1/6.7+..1/98.99+1/100)<5/6 
do -(1/4.5+1/6.7+..1/98.99+1/100)<0 
+A=1/(1.2)+1/(3.4)+...+1/(99.100) 
=1/2+1/12+1/(5.6)+...+1/(99.100) 
=7/12+[1/(5.6)+...1/(99.100)] 
>7/12 do [1/(5.6)+...1/(99.100)]>0

1-1/100=99/100

7/12<2/3<5/6