Cho \(n\in N\) sao, \(n\ge0\). C/m \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)
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Tính tổng đại số
\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)
Tìm n ϵ Z sao cho n là số nguyên
\(\dfrac{2n-1}{n-1};\dfrac{3n+5}{n+1};\dfrac{4n-2}{n+3};\dfrac{6n-4}{3n+4};\dfrac{n+3}{2n-1};\dfrac{6n-4}{3n-2};\dfrac{2n+3}{3n-1};\dfrac{4n+3}{3n+2}\)
1. Cho Ndfrac{1}{31}+dfrac{1}{32}+...+dfrac{1}{60}CMR dfrac{3}{5} N dfrac{4}{5}2. Cho Mdfrac{1}{3}-dfrac{2}{3^2}+dfrac{3}{3^3}-dfrac{4}{3^4}+...+dfrac{29}{3^{29}}-dfrac{30}{3^{30}}CMR M dfrac{3}{16}3. Cho Qdfrac{2}{3}+dfrac{8}{9}+dfrac{26}{27}+...+dfrac{3^{2021}-1}{3^{2021}}CMR Qdfrac{4041}{2}
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1. Cho N=\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}\)
CMR \(\dfrac{3}{5}< N< \dfrac{4}{5}\)
2. Cho M=\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{29}{3^{29}}-\dfrac{30}{3^{30}}\)
CMR \(M< \dfrac{3}{16}\)
3. Cho Q=\(\dfrac{2}{3}+\dfrac{8}{9}+\dfrac{26}{27}+...+\dfrac{3^{2021}-1}{3^{2021}}\)
CMR \(Q>\dfrac{4041}{2}\)
22 tháng 2 2022 lúc 16:48
a, Tính: M = \(1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\)
b, Chứng tỏ: S = \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(n\in N,n\ge2\right)\)
a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)
b: 
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Chứng minh:
Adfrac{1}{1.2.3}+dfrac{1}{2.3.4}+dfrac{1}{3.4.5}+...+dfrac{1}{18.19.20} dfrac{1}{4}
Bdfrac{36}{1.3.5}+dfrac{36}{5.7.9}+dfrac{36}{9.11.13}+...+dfrac{36}{25.27.29} 3
Cdfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{n^2}in 1left(nin N,nge2right)
Ddfrac{1}{4^2}+dfrac{1}{6^2}+dfrac{1}{8^2}+...+dfrac{1}{left(2nright)^2} 4left(nin N,nge2right)
Edfrac{2!}{3!}+dfrac{2!}{4!}+dfrac{2!}{5!}+...+dfrac{2!}{n!} 1left(nin N,nge3right)
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Chứng minh:
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
\(B=\dfrac{36}{1.3.5}+\dfrac{36}{5.7.9}+\dfrac{36}{9.11.13}+...+\dfrac{36}{25.27.29}< 3\)
\(C=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\in< 1\left(n\in N,n\ge2\right)\)
\(D=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< 4\left(n\in N,n\ge2\right)\)
\(E=\dfrac{2!}{3!}+\dfrac{2!}{4!}+\dfrac{2!}{5!}+...+\dfrac{2!}{n!}< 1\left(n\in N,n\ge3\right)\)
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{18.19.20}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.19.20}< \dfrac{1}{4}\)
Cái B TT nhé
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}< 1\)
D TT
E mk thấy nó ss ớ
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1 ,CMR với n \(\in N\),n\(\ge2\).Ta có :
\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+........+\dfrac{1}{n^3}< \dfrac{1}{4}\)
2 , \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^2}+.......+\dfrac{100}{2^{100}}< 2\)
3, CM : \(\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+....+\dfrac{1}{3n+1}< 2\)
Nguyễn Trần Thành ĐạtXuân Tuấn TrịnhHung nguyenHoang HungQuan Ace Legona giúp với
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Chứng minh với mọi \(n\in N;n>1\) Ta có:
\(A=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+.........+\dfrac{1}{n^3}< \dfrac{1}{4}\)
Ta có:\(\dfrac{1}{2^3}< \dfrac{1}{1.2.3};\dfrac{1}{3^3}< \dfrac{1}{2.3.4};\dfrac{1}{4^3}< \dfrac{1}{3.4.5};...;\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)Vậy:\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)Ta có:\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)
=\(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n.\left(n+1\right)}\right)\)=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{n.\left(n+1\right)}\right)\)
=\(\dfrac{1}{4}-\dfrac{1}{2n.\left(n+1\right)}< \dfrac{1}{4}\)
Vì:\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n.\left(n+1\right)}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) hay \(A< \dfrac{1}{4}\)
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Câu 1: Tìm a để dfrac{5a-17}{4a-23} có giá trị lớn nhất.
Câu 2: Cho dfrac{m}{n}1+dfrac{1}{2}+dfrac{1}{3}+...+dfrac{1}{1998} ; m, n in N . CMR m ⋮ 1999
Câu 3: CMR Adfrac{1}{101}+dfrac{1}{102}+dfrac{1}{103}+...+dfrac{1}{200}dfrac{5}{8}
Câu 4: CMR Adfrac{1}{5^2}+dfrac{2}{5^3}+dfrac{3}{5^4}+...+dfrac{n}{5^{n+1}}+...+dfrac{11}{5^{12}} dfrac{1}{16} với n là STN.
Giúp mk với !
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Câu 1: Tìm a để \(\dfrac{5a-17}{4a-23}\) có giá trị lớn nhất.
Câu 2: Cho \(\dfrac{m}{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1998}\) ; m, n \(\in N\) . CMR m \(⋮\) 1999
Câu 3: CMR \(A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}>\dfrac{5}{8}\)
Câu 4: CMR \(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{n}{5^{n+1}}+...+\dfrac{11}{5^{12}}< \dfrac{1}{16}\) với n là STN.
Giúp mk với !
cau 1
de a dat gia tri lon nhat suy ra5a-17/4a-23 lon nhat
suy ra 4a-23 phai nho nhat khac 0 va la so nguyen duong
suy ra 4a-23=1
suy ra 4a=1+23=24
suy ra a=24 chia 4=6
vay de a nho nhat thi a=6
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Chứng minh: Adfrac{2^3+1}{2^3-1}.dfrac{3^3+1}{3^3-1}.dfrac{4^3+1}{4^3-1}....dfrac{9^3+1}{9^3-1} dfrac{3}{2}
Bdfrac{1}{2!}+dfrac{1}{3!}+dfrac{1}{4!}+....+dfrac{1}{n!} 1
Cdfrac{1}{2!}+dfrac{2}{3!}+dfrac{3}{4!}+....+dfrac{n-1}{n!} 1
Dleft(1-dfrac{2}{6}right)left(1-dfrac{2}{12}right)left(1-dfrac{2}{20}right)....left(1-dfrac{2}{nleft(n+1right)}right)dfrac{1}{3}
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Chứng minh: \(A=\dfrac{2^3+1}{2^3-1}.\dfrac{3^3+1}{3^3-1}.\dfrac{4^3+1}{4^3-1}....\dfrac{9^3+1}{9^3-1}< \dfrac{3}{2}\)
\(B=\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+....+\dfrac{1}{n!}< 1\)
\(C=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+....+\dfrac{n-1}{n!}< 1\)
D=\(\left(1-\dfrac{2}{6}\right)\left(1-\dfrac{2}{12}\right)\left(1-\dfrac{2}{20}\right)....\left(1-\dfrac{2}{n\left(n+1\right)}\right)>\dfrac{1}{3}\)