Giúp mình với:
M=\(\frac{1}{1^2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{9^2}\)+\(\frac{1}{10^2}\). Hãy so sánh M với\(\frac{4}{3}\).
Hãy giúp tôi giải phép tính này:
So sánh:
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9^2}+\frac{1}{10^2}+\frac{1}{11^2}\)với\(\frac{5}{12}\)
Ta có
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{11^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)
Mà
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}=\frac{3-2}{2.3}+\frac{4-3}{3.4}\frac{5-4}{4.5}+...+\frac{12-11}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)
Nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{5}{12}\)
1/2^2+1/3^2+1/4^2+....+1/11^2<1/(2.3)+1/(3.4)+1/(4.5)+.....+1/(11.12)
=1/2-13+1/3-1/4+1/5+.....+1/11-1/12
=1/2-1/12=5/12
VẬY A<5/12
ks cho mình nhé
Cho \(M=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+.............+\frac{8}{9!}+\frac{9}{10!}\) . So sánh M với 1 ( với n! = 1.2.3.4..........(n-1).n ; \(n\in\) N*
giúp mình với
\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}\).Hãy so sánh M với \(1\frac{1}{3}\)
Cho \(M=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}+\frac{6}{7}+\frac{7}{8}+\frac{8}{9}+\frac{9}{10}\)
So sánh M với 1
Ta có:
1 = \(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+............+\frac{1}{10}\)(10 phân số \(\frac{1}{10}\))
Mà \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};............;\frac{9}{10}>10\)
\(\Rightarrow M>1\)
Vậy M > 1
Ta có:
1/2=0,5
2/3>0,6
<=>1/2+2/3>1,1>1
<=>1/2+2/3+3/4+...+9/10>1
Vì 1 = \(\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)
\(\Rightarrow\)M > 1 vì \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};...;\frac{9}{10}>\frac{1}{10}\)
\(\Rightarrow M>1\)
1:
Biết M= \(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+...+\frac{1}{2^{43}}-\frac{1}{2^{46}}+\frac{1}{2^{49}}-\frac{1}{2^{52}}\)
Hãy so sánh M và \(\frac{9}{4}\)
\(M=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+....+\frac{1}{2^{43}}-\frac{1}{2^{46}}+\frac{1}{2^{49}}-\frac{1}{2^{52}}\)
Nên \(2^3.M=4-\frac{1}{2}+\frac{1}{2^4}-\frac{1}{2^7}+.....+\frac{1}{2^{46}}-\frac{1}{2^{52}}\)
Suy ra \(2^3.M-M=4-\frac{1}{2^{52}}\)hay\(7.M=4-\frac{1}{2^{52}}\).
Khi đó \(M=\frac{4}{7}-\frac{1}{2^{52}.7}< 1\)
Vì \(\frac{9}{4}>1;M< 1\)nên \(\frac{9}{4}>M\)
Vậy \(\frac{9}{4}>M\)
So sánh A với 1.
Biết: \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{8}{9!}+\frac{9}{10!}\)
Cho M =\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\)
Hãy so sánh với 1
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 = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
B = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
a) So sánh A và B
b) Chứng minh A = \(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)