Tính tỉ số A/B biết: A=1/22.32+1/3.3+...+1/1073.2003
B=1/2.1974+1/3.1975+...+1/31.2003
Những câu hỏi liên quan
Tính P/Q biết:
P = 1/2.32 + 1/3.33 + ... + 1/n.(n+30) + ... + 1/1973.2003
Q = 1/2.1974 + 1/3.1975 + ... + 1/n.(n+1972) + ... + 1/31.2003
\(P=...\)
\(=\frac{1}{30}\left(\frac{30}{2.32}+\frac{30}{3.33}+...+\frac{30}{1973.2003}\right)\)
\(=\frac{1}{30}\left(\frac{1}{2}-\frac{1}{32}+\frac{1}{3}-\frac{1}{33}+...+\frac{1}{1973}-\frac{1}{2003}\right)\)
\(=\frac{1}{30}\left[\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1973}\right)-\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{2003}\right)\right]\)
\(=\frac{1}{30}\left[\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}\right)-\left(\frac{1}{1974}+\frac{1}{1975}+...+\frac{1}{2003}\right)\right]\)
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\(Q=...\)
\(=\frac{1}{1972}\left(\frac{1972}{2.1974}+\frac{1972}{3.1975}+...+\frac{1}{31.2003}\right)\)
\(=\frac{1}{1972}\left(\frac{1}{2}-\frac{1}{1974}+\frac{1}{3}-\frac{1}{1975}+...+\frac{1}{31}-\frac{1}{2003}\right)\)
\(=\frac{1}{1972}\left[\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}\right)-\left(\frac{1}{1974}+\frac{1}{1975}+...+\frac{1}{2003}\right)\right]\)
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Gọi \(\left[\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}\right)-\left(\frac{1}{1974}+\frac{1}{1975}+...+\frac{1}{2003}\right)\right]=A\)
Ta có:\(\frac{P}{Q}=\left(\frac{1}{30}.A\right):\left(\frac{1}{1972}.A\right)=\frac{A}{30}\cdot\frac{1972}{A}=\frac{1972}{30}=\frac{986}{15}\)
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Tính \(\frac{A}{B}\)biết :
\(A=\frac{1}{2.32}+\frac{1}{3.33}+....+\frac{1}{n\left(n+30\right)}+....+\frac{1}{1973.2003}\)
\(B=\frac{1}{2.1974}+\frac{1}{3.1975}+....+\frac{1}{n\left(n+1972\right)}+....+\frac{1}{31.2003}\)
\(A=\frac{1}{2.32}+\frac{1}{3.33}+...+\frac{1}{1973.2003}\)
\(=\frac{1}{30}\left(\frac{1}{2}-\frac{1}{32}+\frac{1}{3}-\frac{1}{33}+...+\frac{1}{1973}-\frac{1}{2003}\right)\)
\(=\frac{1}{30}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1973}-\frac{1}{32}-\frac{1}{33}-\frac{1}{2003}\right)\)
\(=\frac{1}{30}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}-\frac{1}{1974}-\frac{1}{1975}-...-\frac{1}{2003}\right)\)
\(B=\frac{1}{2.1974}+\frac{1}{3.1975}+...+\frac{1}{31.2003}\)
\(=\frac{1}{1972}\left(\frac{1}{2}-\frac{1}{1974}+\frac{1}{3}-\frac{1}{1975}+...+\frac{1}{31}-\frac{1}{2003}\right)\)
\(=\frac{1}{1972}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}-\frac{1}{1974}-\frac{1}{1975}-...-\frac{1}{2003}\right)\)
Vậy \(\frac{A}{B}=\frac{1972}{30}\)
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A= \(\frac{1}{2.32}+\frac{1}{3.33}+...+\frac{1}{\eta.\left(\eta+30\right)}+...+\frac{1}{1973.2003}\)
B= \(\frac{1}{2.1974}+\frac{1}{3.1975}+...+\frac{1}{\eta.\left(\eta+1972\right)}+...+\frac{1}{31.2003}\)
Biết rằng s=1+2.3+\(3.3^2+...+11.3^{10}\)=a+\(\dfrac{21.3^b}{4}\), với a là số hữu tỉ, b là số nguyên. Tính \(P=a+\dfrac{b}{4}\)
\(S=1.3^0+2.3^1+3.3^2+...+11.3^{10}\)
\(3S=1.3^1+2.3^2+...+11.3^{11}\)
\(\Rightarrow S-3S=1+3^1+3^2+...+3^{10}-11.3^{11}\)
\(\Rightarrow-2S=1.\dfrac{3^{11}-1}{3-1}-11.3^{11}\)
\(\Rightarrow-2S=\dfrac{1}{2}.3^{11}-\dfrac{1}{2}-11.3^{11}\)
\(\Rightarrow-2S=-\dfrac{21.3^{11}+1}{2}\)
\(\Rightarrow S=\dfrac{1}{4}+\dfrac{21.3^{11}}{4}\)
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Chứng minh rằng: a, 1/12.22+5/22.32+5/32.42+...+5/92.102 <1 b,1/3+2/32+3/33+...+100/3100 <3/4
Đây Là Lớp Mấy
Tính A-B=...biết A=1.2+2.3+3.4+...+98.99;và B=1+2.2+3.3+...+98.98
tính : A= 1/2.2+1/3.3+1/4.4+...+1/9.9
B=2/3.3+2/5.5+2/7.7+...+2/2007.2007
C=1/4.4+1/6.6+1/8.8+...+1/2006.2006
Tính tỉ số A/B biết:
A=1/1.300+1/2.301+1/3.302+...+1/101.400 và
B=1/1.102+1/2.103+1/3.104+...+1/299.400
a) Tìm số nguyên dương x biết \(3.3^2.3^3.3^4...3^x\)=\(\left(3^3\right)^{12}\)
b) So sánh A= \(\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2017}}\)với \(\frac{1}{4}\)
ta có
\(3^{1+2+3+..+x}=3^{3.12}\Leftrightarrow\frac{x\left(x+1\right)}{2}=36\)
\(\Leftrightarrow x.\left(x+1\right)=72=8.9\Leftrightarrow x=8\)
b. ta có
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+..+\frac{1}{5^{2016}}=\left(\frac{1}{5}+\frac{1}{5^2}+..+\frac{1}{5^{2016}}+\frac{1}{5^{2017}}\right)+1-\frac{1}{5^{2017}}\)
\(=A+1-\frac{1}{5^{2017}}\Rightarrow4A=1-\frac{1}{5^{2017}}< 1\Rightarrow A< \frac{1}{4}\)