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\(S=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(S=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(S=1-\frac{1}{100!}< 1\)
Vậy S<1
thánh đây rồi , đơn giản vậy em nghĩ mãi k ra , cảm ơn anh nhiều
a)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}\)
P/s: Cj chỉ biết làm ý a thôi nhé! Có j ko hiểu cmt nhé!
1.
\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}\)
\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{100-1}{100!}\)
\(=\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}-\dfrac{1}{4!}+...+\)\(\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(=1-\dfrac{1}{100!}< 1\)
2.
\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\)\(\dfrac{1}{100!}\)
Ta có:
\(=\dfrac{1.2}{2!}-\dfrac{1}{2!}+\dfrac{2.3}{3!}-\dfrac{1}{3!}+\dfrac{3.4}{4!}-\dfrac{1}{4!}+...+\)\(\dfrac{99.100}{100!}-\dfrac{1}{100}\)
\(=\left(\dfrac{1.2}{2!}+\dfrac{2.3}{3!}+\dfrac{3.4}{4!}+...+\dfrac{99.100}{100!}\right)\)\(-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)
\(=\left(1+1+\dfrac{1}{2!}+...+\dfrac{1}{98!}\right)\)\(-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)
\(=2-\dfrac{1}{99!}-\dfrac{1}{100!}< 2\)
a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)
\(\Rightarrowđpcm\)
d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\Rightarrowđpcm\)
Ta có:
\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+...+\dfrac{99.100-1}{100!}\)
\(=\dfrac{1.2}{2!}-\dfrac{1}{2!}+\dfrac{2.3}{3!}-\dfrac{1}{3!}+...+\dfrac{99.100}{100!}-\dfrac{1}{100!}\)
\(=\left(\dfrac{1.2}{2!}+\dfrac{2.3}{3!}+...+\dfrac{99.100}{100!}\right)-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)
\(=\left(1+1+\dfrac{1}{2!}+...+\dfrac{1}{98!}\right)-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)
\(=1+1-\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(=2-\dfrac{1}{99!}-\dfrac{1}{100!}< 2\)
Vậy \(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+...+\dfrac{99.100-1}{100!}< 2\) (Đpcm)
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+..+\dfrac{1}{9900}\)
\(A=\left(\dfrac{1}{2}+\dfrac{1}{12}\right)+\left(\dfrac{1}{30}+...+\dfrac{1}{9900}\right)\)
\(A>\dfrac{1}{2}+\dfrac{1}{12}\Rightarrow A>\dfrac{7}{12}\left(1\right)\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{5}{6}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< \dfrac{5}{6}\left(2\right)\)
\(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\rightarrowđpcm\)
Ta có :
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+..........+\dfrac{1}{99.100}\)
\(\Leftrightarrow A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+............+\dfrac{1}{99.100}>\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{7}{12}\)
\(\Leftrightarrow A>\dfrac{1}{12}\)\(\left(1\right)\)
Lại có :
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...........+\dfrac{1}{99.100}\)
\(\Leftrightarrow A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-.........-\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
\(\Leftrightarrow A< \dfrac{5}{6}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\rightarrowđpcm\)
Ta có : \(B\text{=}\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{99.100}\)
\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{100}\)
\(B\text{=}\dfrac{247}{300}\)
Ta có : \(\dfrac{7}{12}\text{=}\dfrac{175}{300};\dfrac{5}{6}\text{=}\dfrac{250}{300}\)
Vì : \(\dfrac{175}{300}< \dfrac{247}{300}< \dfrac{250}{300}\)
\(\Rightarrowđpcm\)
\(S=\dfrac{1}{1.2}+\dfrac{2}{1.2.3}+........+\dfrac{99}{1.2.......100}\)
\(=\dfrac{1}{2!}+\dfrac{2}{3!}+....+\dfrac{99}{100!}\)
\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+.......+\dfrac{100-1}{100!}\)
\(=\dfrac{1}{1}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+....+\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(=1-\dfrac{1}{100!}< 1\)
\(\Leftrightarrow S< 1\left(đpcm\right)\)