Tính :
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(C=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
Gửi
TNs tao cuồng:c/m \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{3}{2^3}+....+\frac{100}{2^{100}}<2\)Ta có:\(2B=1+\frac{1}{2}+\frac{3}{2^2}+....+\frac{100}{2^{99}}\)\(\Rightarrow2B-B=B=1+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)(*)c/m \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}<1\)Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)\)\(\Rightarrow A=1-\frac{1}{2^{99}}<1\)do đó \(B=1+A-\frac{100}{2^{100}}\Rightarrow B<2-\frac{100}{2^{100}}<2\left(đpcm\right)\)
Tính tổng
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^{^3}}+...+\frac{1}{2^{100}}\)
\(C=\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^{3.}}+.............+\frac{1}{2^{100}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+.................+\frac{1}{2^{99}}\)
\(2B-B=1-\frac{1}{2^{100}}\)
\(B=1-\frac{1}{2^{100}}\)
\( C=\frac{1}{2}-\frac{1}{2^2}+.................+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(2 C=1-\frac{1}{2}+......................+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)
\(2 C+C=1-\frac{1}{2^{100}}\)
\(C=\left(1-\frac{1}{2^{100}}\right):3\)
Tính Q=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{100}}{\frac{100-1}{1}+\frac{102-2}{2}+...+\frac{100-99}{99}}\)
Sửa đề:
\(Q=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{\frac{100-1}{1}+\frac{100-2}{2}+...+\frac{100-99}{99}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{100-1+\frac{100}{2}-1+...+\frac{100}{99}-1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{\frac{100}{100}+\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{100.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}=\frac{1}{100}\)
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 \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
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. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
Tính:
\(E=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(F=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(F=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(F=\left(\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
\(F=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)-2.\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
\(F=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{50}}\right)\)
\(F=\frac{1}{2^{51}}+\frac{1}{2^{52}}+...+\frac{1}{2^{100}}\)
\(E=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2E=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2E-E=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(E=1-\frac{1}{2^{100}}\)
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
Tính dãy số sau :
\(D=\frac{100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{98}{99}+\frac{99}{100}}\)
a, Tính : \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b, Tính : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c, Tính : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)