(1+1/2)(1+1/3)...(1+1/n)
=3/2.4/3....n+1/n
=(3.4....n+1)/(2.3....n)=n+1/2
1/2+1/3+1/4+..+1/n<1/2.(n-1)=(n-1)/2 (do có n-1 số hạng) <n+1/2
=> lớn hơn
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 \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
CMR \(\forall n\in\)N* ta có
\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+...+\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\)
Tính giá trị của biểu thức:
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)
\(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)
\(S=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
\(B=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{100}\right)\)
\(D=\left(1-\frac{1}{17}\right)\left(1-\frac{2}{17}\right)\left(1-\frac{3}{17}\right)...\left(1-\frac{27}{17}\right)\)
\(D=\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{25}\right)\right]:\left[\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{25}\right)\right]\)
\(H=\frac{\left(1+97\right)\left(1+\frac{97}{2}\right)\left(1+\frac{97}{3}\right)\left(1+\frac{97}{4}\right)+...+\left(1+\frac{97}{99}\right)}{\left(1+99\right)\left(1+\frac{99}{2}\right)\left(1+\frac{99}{3}\right)\left(1+\frac{99}{4}\right)+...+\left(1+\frac{99}{97}\right)}\)
Tính giá trị biểu thức :
1. \(A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
2. \(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
3. \(C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
4. \(D=(\frac {150}{1111}+\frac{5}{75}-\frac{14}{77})(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}) \)
5. Cho \(M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right);N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\). Tính \(P=M-N\)
6. \(E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
7. \(F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
8. \(G=\left[\frac{\left(6-4\frac{1}{2}\right):0,03}{\left(3\frac{1}{20}-2,65\right).4+\frac{2}{5}}-\frac{\left(0,3-\frac{3}{20}\right).1\frac{1}{2}}{\left(1,88+2\frac{3}{25}\right).\frac{1}{80}}\right]:\frac{49}{60}\)
9. \(H=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{4.5.6}+...+\frac{1}{98.99.100}\)
10. \(I=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2499}{2500}\)
11. \(k=\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{999}\right)\)
12. \(L=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}...\)(98 thừa số)
13. \(M=-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{3}}}}\)
14. \(N=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}\)
15. \(P=\left(\frac{1}{4}-1\right)\left(\frac{1}{5}-1\right)...\left(\frac{1}{2000}-1\right)\left(\frac{1}{2001}-1\right)\)
16. \(Q=\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\right):\left(\frac{1}{1004.2006}+\frac{1}{1005.2005}+...+\frac{1}{2006.1004}\right)\)
a)Tìm số nguyên dương n thỏa mãn:
\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)=\frac{2013}{2014}\)
b)tìm a sao cho
\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)=11.a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
Chứng minh rằng:
a)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\)<1
b)\(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)<2
c)\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)<\(\frac{3}{4}\)
d)\(\frac{1}{3^3}+\frac{1}{4^3}+\frac{1}{5^3}+...+\frac{1}{n^3}\)<\(\frac{1}{12}\)\(\left(n\in N;n\ge3\right)\)
e)\(\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)<1 (n nguyên dương)
g)\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{2048}\)>3
h)\(\left(\frac{2}{1}\right)\left(\frac{4}{3}\right)\left(\frac{6}{5}\right)...\left(\frac{200}{199}\right)\)
Cho M = \(\frac{\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right)}{\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)}\);
N = \(\frac{\left(92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}\right)}{\left(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}\right)}\)
Tìm tỉ số phần trăm của M và N
