Bài 1: Cho B=1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}+...+\dfrac{1}{2^{100}-1}\)
a) Chứng minh B<100
b) Chứng minh B>50
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.....+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) Chứng minh A < \(\dfrac{3}{16}\)
2/ Cho B=(\(\dfrac{1}{2^2}\)-1)(\(\dfrac{1}{3^2}\)-1)....(\(\dfrac{1}{100^2}\)-1) So sánh B và \(\dfrac{-1}{2}\)
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
Bài 7:
Chứng minh rằng: \(\dfrac{3}{10}< \dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}.\)
\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)
\(\dfrac{1}{4^2}>\dfrac{1}{4\cdot5}=\dfrac{1}{4}-\dfrac{1}{5}\)
...
\(\dfrac{1}{100^2}>\dfrac{1}{100}-\dfrac{1}{101}\)
Do đó: \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}>\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}>\dfrac{90.9}{303}=\dfrac{3}{10}\)(1)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3}-\dfrac{1}{4}\)
...
\(\dfrac{1}{100^2}< \dfrac{1}{99}-\dfrac{1}{100}\)
Do đó: \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=>\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}< \dfrac{50}{100}=\dfrac{1}{2}\)(2)
Từ (1),(2) suy ra \(\dfrac{3}{10}< \dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
Cho A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+....+\(\dfrac{1}{100^2}\). Chứng minh rằng A<\(\dfrac{3}{4}\)
Cho A=1+\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2^{100}-1}\)
Chứng minh rằng 50<A<100
Cho: A= 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2^{100}-1}\)
Chứng minh rằng: 50 < A < 100
Giúp mình với!
\(\text{Bài 4. Chứng tỏ rằng:}\)
\(a\)) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}< 1\)
\(b\)) \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(c\)) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
\(d\)) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}< 1\)
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
Chứng Minh Rằng : A= \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{100^2}\) <\(\dfrac{3}{4}\)
Lời giải:
$A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{99.100}$
$A< \frac{1}{4}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$
$A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$
$A< \frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{1}{4}+\frac{1}{2}$
Hay $A< \frac{3}{4}$
a , cho A = \(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}\) . Chứng minh A < \(\dfrac{7}{4}\)
b ,cho B = 21 + 22 + 23 + ... + 260 . Chứng minh B \(⋮\) 21
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
Chứng tỏ rằng:
a)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{3}{4}\)
b)\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
b\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4
Tương tự như vậy với câu a\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2