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Ta có:\(A< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Mặt khác:\(A>\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)
Vậy \(\frac{8}{9}>A>\frac{2}{5}\)
Đặt \(A=5+5^2+5^3+...+5^{1002}\)
- Chứng tỏ chia hết cho 6
\(A=5+5^2+5^3+...+5^{1002}\)
\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{1001}+5^{1002}\right)\)
\(A=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{1001}\left(1+5\right)\)
\(A=5.6+5^3.6+...+5^{1001}.6\)
\(A=6\left(5+5^3+...+5^{1002}\right)\) chia hết cho 6(đpcm)
Các bài sau làm tương tự
Ta có :
\(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...........+\dfrac{1}{19}\)
\(B=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+......+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+..........+\dfrac{1}{19}\right)\)
Ta thấy :
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}>\dfrac{1}{9}+\dfrac{1}{9}+\dfrac{1}{9}+\dfrac{1}{9}+\dfrac{1}{9}\)\(=\dfrac{1}{9}.5=\dfrac{5}{9}\)\(>\dfrac{1}{2}\)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+......+\dfrac{1}{19}>\dfrac{1}{19}+\dfrac{1}{19}+.........+\dfrac{1}{19}\)\(=\dfrac{1}{19}.5=\dfrac{5}{19}\)\(>\dfrac{1}{2}\)
\(\Rightarrow B>\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{5}{4}>1\)
\(\Rightarrow B>1\rightarrowđpcm\)
~ Chúc bn học tốt ~
\(M=1+5+5^2+...+5^{2005}\)
\(\Rightarrow5M=5+5^2+5^3+...+5+5^{2006}\)
\(\Rightarrow5M-M=\left(5+5^2+...+5^{2006}\right)-\left(1+5+...+5^{2005}\right)\)
\(\Rightarrow5M-M=4M=5^{2006}-1\Rightarrow M=\frac{5^{2006}-1}{4}\)
\(\frac{N}{4}=\frac{5^{2006}}{4}>\frac{5^{2006}-1}{4}=M\Rightarrow M< \frac{N}{4}\)
a) S=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)
2S=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2017.2019}\)
2S=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)
2S=\(1-\dfrac{1}{2019}\)
2S=\(\dfrac{2018}{2019}\)
S\(\dfrac{1009}{2019}\)
\(A=4+2^2+2^3+...+2^{2005}\)
\(2A=4+2^2+2^3+...+2^{2006}\)
\(2A-A=\left(4+2^2+2^3+...+2^{2006}\right)-\left(4+2^2+2^3+...+2^{2005}\right)\)
\(A=4+2^2+2^3+...+2^{2006}-4-2^2-2^3-...-2^{2005}\)
\(A=2^{2006}\)
Vậy A là 1 luỹ thừa của cơ số 2
\(B=5+5^2+...+5^{2021}\)
\(5B=5^2+5^3+...+5^{2022}\)
\(5B-B=\left(5^2+5^3+...+5^{2022}\right)-\left(5+5^2+...+5^{2021}\right)\)
\(4B=5^{2022}-5\)
\(B=\frac{5^{2022}-5}{4}\)
\(B+8=\frac{5^{2022}-5}{4}+8\)
\(B+8=\frac{5^{2022}-5}{4}+\frac{32}{4}\)
\(B+8=\frac{5^{2022}-5+32}{4}\)
\(B+8=\frac{5^{2022}+27}{4}\)
=> B + 8 k thể là số b/ph của 1 số tn
\(A=5+5^2+...+5^{2005}\)
\(5A=5^2+5^3+...+5^{2006}\)
\(5A-A=\)\(\left(5^2+5^3+...+5^{2006}\right)-\left(5+5^2+...+5^{2005}\right)\)
\(A=\frac{5^{2006}-5}{4}\)