CMR:
\(\dfrac{1}{5^3}+\dfrac{1}{6^3}+\dfrac{1}{7^3}+...+\dfrac{1}{2004^3}< \dfrac{1}{40}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
29 tháng 3 2023
A=1/5^3+1/6^3+...+1/2023^3
1/5^3<1/4*5*6
Xét tương tự, ta đều sẽ được:
\(\dfrac{1}{n^3}< \dfrac{1}{n\left(n-1\right)\left(n+1\right)}\)
=>\(A< \dfrac{1}{4\cdot5\cdot6}+\dfrac{1}{5\cdot6\cdot7}+...+\dfrac{1}{2022\cdot2023\cdot2024}\)
=>\(A< \dfrac{1}{2}\left(\dfrac{2}{4\cdot5\cdot6}+\dfrac{2}{5\cdot6\cdot7}+...+\dfrac{2}{2022\cdot2023\cdot2024}\right)\)
=>\(A< \dfrac{1}{2}\left(\dfrac{1}{4\cdot5}-\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}-\dfrac{1}{7\cdot8}+...+\dfrac{1}{2022\cdot2023}-\dfrac{1}{2023\cdot2024}\right)\)
=>A<1/40
Ta có BĐT: \(\dfrac{1}{k\left(k+1\right)\left(k+2\right)}< \dfrac{1}{k^3}< \dfrac{1}{\left(k-1\right)\cdot k\cdot\left(k+1\right)}\)
Do đó, ta được:
\(\dfrac{1}{5\cdot6\cdot7}+\dfrac{1}{6\cdot7\cdot8}+...+\dfrac{1}{2023\cdot2024\cdot2025}< A\)
\(\Leftrightarrow A>\dfrac{1}{2}\left(\dfrac{1}{5\cdot6}-\dfrac{1}{2024\cdot2025}\right)>\dfrac{1}{2}\left(\dfrac{1}{30}-\dfrac{1}{390}\right)=\dfrac{1}{65}\)
=>1/65<A<1/40
NN
15 tháng 11 2017
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
15 tháng 11 2017
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
31 tháng 10 2022
6:
\(4D=2^2+2^4+...+2^{202}\)
=>3D=2^202-1
hay \(D=\dfrac{2^{202}-1}{3}\)
7: \(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{32}{99}=\dfrac{16}{99}\)
5 tháng 7 2017
chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D
Ta có: \(n^3-n< n^3\forall n\)
mà: \(n^3-n=n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)
Nên: \(\left(n-1\right)n\left(n+1\right)< n^3\Leftrightarrow\dfrac{1}{\left(n-1\right)n\left(n+1\right)}>\dfrac{1}{n^3}\)
Trở lại bài toán:
\(SV=\dfrac{1}{5^3}+\dfrac{1}{6^3}+\dfrac{1}{7^3}+...+\dfrac{1}{2004^3}< \dfrac{1}{\left(5-1\right).5.\left(5+1\right)}+\dfrac{1}{\left(6-1\right).6.\left(6+1\right)}+\dfrac{1}{\left(7-1\right).7.\left(7+1\right)}+...+\dfrac{1}{\left(2004-1\right).2004.\left(2004+1\right)}\)
\(SV< \dfrac{1}{4.5.6}+\dfrac{1}{5.6.7}+\dfrac{1}{6.7.8}+...+\dfrac{1}{2003.2004.2005}=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}+\dfrac{1}{5.6}-\dfrac{1}{6.7}+\dfrac{1}{6.7}-\dfrac{1}{7.8}+...+\dfrac{1}{2003.2004}-\dfrac{1}{2004.2005}\right)=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{2004.2005}\right)=\dfrac{1}{2.4.5}-\dfrac{1}{2.2004.2005}=\dfrac{1}{40}-\dfrac{1}{2.2004.2005}< \dfrac{1}{40}\left(đpcm\right)\)