So sánh với 3
\(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+....+\left(1+\frac{2}{n^2+3n}\right)\)
A=\(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{2}{n^2+3n}\right)\)
So sánh A với 3
Với n =1 thì A < 3. Vậy ta phải đi chứng minh A < 3
Giả sử A < 3 đúng với n = k. Ta có:
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{2}{k^2+3k}\right)< 3\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(\frac{k^2+3k+2}{k\left(k+3\right)}\right)\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}\)
Ta phải đi chứng minh A < 3 đúng với n = k +1 tức là phải chứng minh:
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}+\left(1+\frac{2}{\left(k+1\right)^2+3\left(k+1\right)}\right)\) \(< 3+\frac{\left(k+2\right)\left(k+3\right)}{\left(k+1\right)\left(k+4\right)}\)
Ta sẽ có:
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}+\left(1+\frac{2}{k^2+2k+1+3k+3}\right)\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}+\frac{k^2+5k+6}{k^2+5k+4}\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}+\frac{\left(k+2\right)\left(k+3\right)}{\left(k+1\right)\left(k+4\right)}\) \(< 3+\frac{\left(k+2\right)\left(k+3\right)}{\left(k+1\right)\left(k+4\right)}\)
Vậy A đúng với n = k + 1 thì A đúng với n = k
Vậy A < 3 là điều phải chứng minh.
(Phương pháp quy nạp toán học)
so sánh với 3
các bạn giúp mik với
\(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{2}{n^2+3n}\right)\)
\(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{2}{n^2+3n}\right)\)
\(=\left(1+1+1\right)+\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{9}+...+\frac{2}{n^2+3n}\right)+\left(1+1+1+...+1\right)\)
\(=3+\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{9}+...+\frac{2}{n^2+3n}\right)+\left(1+1+1+...+1\right)\)
Có: \(\frac{1}{2}+\frac{1}{5}+\frac{1}{9}+...+\frac{2}{n^2+3n}>0\)
\(1+1+1+...+1>0\)
=> \(3+\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{9}+...+\frac{2}{n^2+3n}\right)+\left(1+1+1+...+1\right)>3\)
Hay \(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{2}{n^2+3n}\right)>3\)
CMR với n thuộc Z, ta có:
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{5}\right).\left(1+\frac{1}{9}\right)...\left(1+\frac{2}{n^2+3n}\right)< 3\)
Ta thấy: 1+ 2/ n^2+3n = n^2+3n+2 / n(n+3) =(n+1)(n+2) /n(n+3)
Áp dụng công thức trên,ta có:
A= (1+2/4 )(1+ 2/10)(1+2/18).....(1+2/ n^2+3n)
=(1+2 /1x4)( 1+2 /2x5)(1+2 /3x6).....[ (n+1)(n+2)/ n(n+3)]
=(2x3 /1x4)(3x4 /2x5)(4x5 /3x6).....[ (n+1)(n+2) /n(n+3)]
= 3x(n+1 /n+3)
Vì n+1 /n+3 <1 với mọi n thuộc N nên 3x(n+1 /n+3) <3
Vậy A<3
Bài 1 : cho 2 biểu thức
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)
So sánh A với \(\frac{1}{21}\)
So sánh B với \(\frac{11}{21}\)
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2....18.19}{2.3...19.20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Vậy A > 1/21
Cho n ∈ N*. Chứng minh rằng
B = \(\left(1+\frac{1}{2}\right)-\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)...\left(1+\frac{1}{n^3+3n}\right)\) < 3
1. tính A= \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}\)
2. tính B= \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}\)
3. So sánh C= \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)với \(\frac{1}{21}\)
4. So sánh D= \(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{100}\right)\)với \(\frac{11}{19}\)
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
cho A=\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{9}-1\right)\left(\frac{1}{10}-1\right)\)
so sánh A với \(\frac{-1}{9}\)
So sánh các số :
\(\left(\frac{1}{2}\right)^1;\left(\frac{1}{3}\right)^{-1};\left(\frac{1}{2}\right)^2;\left(\frac{1}{4}\right)^{-1};\left(\frac{1}{3}\right)^{-2}\)
(1/2)^-1=2
(1/2)^-2=4
có 2<4
=>(1/2)^-1<(1/2)^-2
Ta có :
\(\left(\frac{1}{2}\right)^{-1}=\left(2^{-1}\right)^{-1}=2\)
\(\left(\frac{1}{3}\right)^{-1}=3\)
\(\left(\frac{1}{2}\right)^{-2}=\left(2^{-1}\right)^{-2}=2^2=4\)
\(\left(\frac{1}{4}\right)^{-1}=\left(4^{-1}\right)^{-1}=4\)
\(\left(\frac{1}{3}\right)^{-2}=\left(3^{-1}\right)^{-2}=3^2=9\)
Do đó ta có :
\(\left(\frac{1}{2}\right)^{-1}< \left(\frac{1}{3}\right)^{-1}< \left(\frac{1}{2}\right)^{-2}=\left(\frac{1}{4}\right)^{-1}< \left(\frac{1}{3}\right)^{-2}\)
a,\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
b,\(\left[\frac{5}{7}-\frac{7}{5}\right]-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)
c,\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{71}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{8}\)
d,\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
e,\(\left(\frac{1}{2}-\frac{13}{14}\right):\frac{5}{7}-\left(\frac{-2}{21}+\frac{1}{7}\right):\frac{5}{7}\)
g,\(\frac{4}{9}:\left(\frac{-1}{7}\right)+6\frac{5}{9}:\left(\frac{-1}{7}\right)\)