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b,\(D=2.\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{n.\left(n+2\right)}\right)\)
\(\Rightarrow D=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{n.\left(n+2\right)}\)
\(\Rightarrow D=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n.\left(n+2\right)}\)
\(\Rightarrow D=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+2}\)
\(\Rightarrow D=1-\frac{1}{n+2}=\frac{n}{n+2}< \frac{n+2}{n+2}=1\left(1\right)\)
\(\Rightarrow D=\frac{n}{n+2}>0\left(2\right)\)
Từ (1);(2)\(\Rightarrow0< D< 1\)
\(\Rightarrowđpcm\)
a,\(C>0\)
\(C=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{19}< 9;\frac{1}{11}< 1\)
\(\Rightarrow0< A< 1\)
\(\Rightarrow A\notinℤ\)
c,\(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
Ta quy đồng 3 số đầu
\(=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}>\frac{6.2}{12}=1\)
\(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
\(=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}< \frac{6.2}{6}=2\)
\(1< E< 2\)
\(E\notinℤ\)
Ta có : D = \(2\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{25}+.....+\frac{1}{n\left(n+2\right)}\right)\)
\(\Rightarrow D=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{n\left(n+2\right)}\)
\(\Rightarrow D=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{n}-\frac{1}{n+1}\)
\(\Rightarrow D=1-\frac{1}{n+1}=\frac{n+1}{n+1}-\frac{1}{n+1}=\frac{n}{n+1}\)
Vậy D không phải là số nguyên (đpcm)
\(D=2.\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{n\left(n+2\right)}\right)\)
\(D=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{n\left(n+2\right)}\)
\(D=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n\left(n+2\right)}\)
\(D=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{\left(n+2\right)-n}{n\left(n+2\right)}\)
\(D=\frac{3}{1.3}-\frac{1}{1.3}+\frac{5}{3.5}-\frac{3}{3.5}+\frac{7}{5.7}-\frac{5}{5.7}+...+\frac{\left(n+2\right)}{n\left(n+2\right)}-\frac{n}{n\left(n+2\right)}\)
\(D=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\)
\(D=\frac{1}{1}-\frac{1}{n+2}\)
\(D=\frac{n+2}{n+2}-\frac{1}{n+2}\)
\(D=\frac{n+2-1}{n+2}\)
\(D=\frac{n+1}{n+2}\Rightarrow D\notin Z\left(dpcm\right)\)
S=1-1/4+1-1/9+...+1-1/x2
S=(1+1+1+...+1)-(1/4+1/9+...+1/x2)
Có (1/4+1/9+...+1/x2)<1/(1.2)+1/(2.3)+...+1/(x-1)x=1-1/x<1
=> (1/4+1/9+...+1/x2) ko là số nguyên
=>S ko là số nguyên
chịu