CMR:S=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\frac{2}{3}\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+....+\frac{100}{2^{101}}\)\(A-\frac{A}{2}=\left(1+\frac{3}{2^3}+....+\frac{100}{2^{100}}\right)-\left(\frac{1}{2}+\frac{3}{2^4}+.....+\frac{100}{2^{101}}\right)\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{100}}-\frac{1}{2^{101}}\)
\(\frac{A}{2}=\left(1-\left(\frac{1}{2}\right)^{101}\right).2-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{2^{101}-1}{2^{100}}-\frac{100}{2^{101}}\)
\(A=\frac{2^{101}-1}{2^{99}}-\frac{100}{2^{100}}\)
chứng tỏ rằng
C = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}< \frac{1}{3}\)
D = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Phần C đề thiếu
\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)
\(\Rightarrow4D=3-\frac{203}{3^{100}}\)
\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)
\(C=\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(\Rightarrow2C=1-\frac{1}{2}+...+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)
\(\Rightarrow2C+C=(1-\frac{1}{2}+...+\frac{1}{2^{98}}-\frac{1}{2^{99}})+\)\((\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}})\)
\(\Rightarrow3C=1-\frac{1}{100}\)
\(\Rightarrow C=\frac{1}{3}-\frac{1}{300}< \frac{1}{3}\left(đpcm\right)\)
a)\(\frac{1}{10^2}+\frac{1}{11^2}+\frac{1}{12^2}+...+\frac{1}{100^2}<\frac{3}{4}\)
b)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<\frac{99}{100}\)
c)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<\frac{3}{4}\)
a,1/102+1/112+1/122+...+1/1002<1/9.10+1/10.11+1/11.12+...+1/99.100=1/9-1/10+1/10-1/11+...+1/99-1/100
=1/9-1/100=91/900<3/4
Vậy 1/102+1/112+1/122+...+1/1002<3/4
b,1/22+1/32+1/42+...+1/1002<1/1.2+1/2.3+1/3.4+...+1/99.100=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100=99/100
Vậy 1/22+1/32+1/42+...+1/1002<99/100
c,1/22+1/32+1/42+...+1/1002<1/22+(1/2.3+1/3.3+...+1/99.100)=1/4+(1/2-1/3+1/3-1/4+...+1/99-1/100)
=1/4+(1/2-1/100)=1/4+49/100=74/100<3/4=75/100
Vậy 1/22+1/32+1/42+...+1/1002<3/4
Chứng minh rằng:
a/\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
b/\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1\frac{3}{4}\)
c/\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Chứng tỏ rằng
a) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
b) \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}< \frac{3}{4}\)
c) \(\frac{1}{3}+\frac{3}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
CMR : E = \(1-\frac{1}{2^2}-\frac{1}{3^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)
F = \(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{200^2}< \frac{1}{2}\)
H = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
\(E=1-\frac{1}{2^2}-\frac{1}{3^2}-..........-\frac{1}{2004^2}\)
\(E=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+..........+\frac{1}{2014^2}\right)\)
Ta có : \(E< 1-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{2003.2004}\right)\\ \)
Đặt A= \(1-\left(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2003.2004}\right)\\ =>A=1-\left(1-\frac{1}{2004}\right)\\ =>A=1-\frac{2003}{2004}\\ =>A=\frac{1}{2004}\)
Chắc chắn bạn đã ghi nhầm dấu
Chứng minh rằng:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\) \(B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}< \frac{3}{4}\)
Giúp mình nha. Bài cuối cùng của đề toán dài 36 bài của mình đó
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)
Nên từ đây => \(A< 1\) (ĐPCM)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{100^2}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)
Cộng vế theo vế
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
=> \(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(A< \frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)(1)
Lại có \(\frac{99}{100}< 1\)(2)
Từ (1) và (2) => \(A< \frac{99}{100}< 1\Rightarrow A< 1\left(đpcm\right)\)