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Bài 2 :
\(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+............+\dfrac{2017}{4^{2017}}\)
\(\Leftrightarrow4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...........+\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+..........+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+..........+\dfrac{2017}{4^{2017}}\right)\)
\(\Leftrightarrow3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+.........+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)
Đặt :
\(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow4A=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2015}}\)
\(\Leftrightarrow4A-A=\left(4+1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2016}}\right)\)
\(\Leftrightarrow3A=4-\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow D=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}\)
\(\Leftrightarrow3S=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}-\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow3S< \dfrac{4}{3}\)
\(\Leftrightarrow S< \dfrac{4}{9}\)
\(\Leftrightarrow S< \dfrac{1}{2}\rightarrowđpcm\)
\(A=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\) ( A cho đẹp :v)
\(4A=4\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)
\(4A=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)
\(4A-A=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)\(3A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)
Đặt:
\(M=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\)
\(4M=4\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)
\(4M=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)
\(4M-M=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3M=4-\dfrac{1}{4^{2016}}\)
\(M=\dfrac{4}{3}-\dfrac{1}{4^{2016}}\)
Thay M vào A ta có:
\(A=\dfrac{4}{9}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)
\(\Rightarrow A< \dfrac{1}{2}\Rightarrowđpcm\)
a) \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{2018}{2019!}\\ =\left(\dfrac{1}{1!}-\dfrac{1}{2!}\right)+\left(\dfrac{1}{2!}-\dfrac{1}{3!}\right)+...+\left(\dfrac{1}{2018!}-\dfrac{1}{2019!}\right)\\ =1-\dfrac{1}{2019!}< 1\)
b) \(\dfrac{1\cdot2-1}{2!}+\dfrac{2\cdot3-1}{3!}+...+\dfrac{999\cdot1000-1}{1000!}\\ =\dfrac{1\cdot2}{2!}-\dfrac{1}{2!}+\dfrac{2\cdot3}{3!}-\dfrac{1}{3!}+...+\dfrac{999-1000}{1000!}-\dfrac{1}{1000!}\\ =\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{1!}-\dfrac{1}{3!}+\dfrac{1}{2!}-\dfrac{1}{4!}+...+\dfrac{1}{999!}+\dfrac{1}{1000!}\\ =1+1-\dfrac{1}{1000!}\\ =2-\dfrac{1}{1000!}< 2\)
a)
\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...........-\dfrac{1}{2004^2}\)
\(\Leftrightarrow B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..............+\dfrac{1}{2004^2}\right)\)
Đặt :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.............+\dfrac{1}{2004^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..........................
\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..............+\dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{2003}-\dfrac{1}{2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2004}\)
\(\Leftrightarrow A< \dfrac{2003}{2004}\)
\(\Leftrightarrow1-A< 1-\dfrac{2003}{2004}\)
\(\Leftrightarrow B< \dfrac{1}{2004}\left(đpcm\right)\)
b) \(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-........+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
\(\Leftrightarrow2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.....+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+....+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(\Leftrightarrow4S=1-\dfrac{1}{2^2}+.......+\dfrac{1}{2^{4n}}-\dfrac{1}{2^{4n+2}}+.......+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\)
\(\Leftrightarrow4S+S=\left(1-\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)\(\Leftrightarrow5S=1-\dfrac{1}{2^{2004}}< 1\)
\(\Leftrightarrow S< \dfrac{1}{5}=0,2\)
\(\Leftrightarrow S< 0,2\left(đpcm\right)\)
cho mik hỏi mik ko hiểu tại sao từ 1/2^4n-2 khi nhân với 2^2 lại ra đc 1/2^4n vậy? Xin hãy giải đáp giùm mik
a) \(P=\frac{1+2}{1^2.2^2}+\frac{2+3}{2^2.3^2}+...+\frac{9+10}{9^2.10^2}\)
\(P=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\) ( rút gọn số mũ nhé )
\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\)
\(P=1-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
Vì \(\frac{9}{10}< 1\Rightarrow P< 1\) (đpcm)
b) Chút nữa mình làm nhé ^^
b)
\(Q=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
Ta so sánh giữa A và Q.
\(\frac{1}{1.2}>\frac{1}{3};\frac{1}{2.3}>\frac{1}{3^2};\frac{1}{3.4}>\frac{1}{3^3};....;\frac{1}{100.101}>\frac{1}{3^{100}}\)
\(\Rightarrow Q< A\)
Ta lại tiếp tục so sánh A và \(\frac{1}{2}\)
Ta có :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\Leftrightarrow A< \frac{1}{2}\)
Ta được:
\(Q< A< \frac{1}{2}\Leftrightarrow Q< \frac{1}{2}\)
\(S=\dfrac{1}{1.2}+\dfrac{2}{1.2.3}+........+\dfrac{99}{1.2.......100}\)
\(=\dfrac{1}{2!}+\dfrac{2}{3!}+....+\dfrac{99}{100!}\)
\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+.......+\dfrac{100-1}{100!}\)
\(=\dfrac{1}{1}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+....+\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(=1-\dfrac{1}{100!}< 1\)
\(\Leftrightarrow S< 1\left(đpcm\right)\)
nhân C vs 3 sau đó lấy 3C-C sẽ ra đc 2 C = 1 - 1/399 => C= 1/2 - 1/ (2x399 )
*chứng minh:
\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+n}< 1\)
Xét thừa số tổng quát:
\(\dfrac{1}{1+2+3+...+n}=\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=\dfrac{2}{n\left(n+1\right)}\)
\(\Rightarrow\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+n}\)
\(=\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{n\left(n+1\right)}=2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)=1-\dfrac{1}{2\left(n+1\right)}< 1\)
Điều phải chứng minh ~
p/s: xem lại,tối t hay ngáo lắm,nghĩ thế này gõ thế khác đó ^^
Èo