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29 tháng 8 2017
A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)
= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)
Vậy ...
11 tháng 3 2017
Dạng tổng quát: \(\sqrt[k+1]{\frac{k+1}{k}}>\sqrt[k+1]{\frac{k+1}{k+1}}=1\) với k = 1; 2; 3; ...; n
=> \(a=\sqrt{2}+\sqrt[3]{\frac{3}{2}}+\sqrt[4]{\frac{4}{3}}+...+\sqrt[n+1]{\frac{n+1}{n}}>n\) (1)
Áp dụng bđt AM-GM cho k + 1 số dương ta có:
\(\sqrt[k+1]{\frac{k+1}{k}}=\sqrt[k+1]{1.1.1...1.\frac{k+1}{k}}< \frac{1+1+1+...+1+\frac{k+1}{k}}{k+1}=\frac{1.k}{k+1}+\frac{\frac{k+1}{k}}{k+1}\)
\(\Leftrightarrow\sqrt[k+1]{\frac{k+1}{k}}< \frac{k}{k+1}+\frac{1}{k}=1-\frac{1}{k+1}+\frac{1}{k}=1+\left(\frac{1}{k}-\frac{1}{k+1}\right)\)
\(< 1+\frac{1}{k\left(k+1\right)}\)
Áp dụng vào bài ta được:
\(a< \left(1+\frac{1}{1.2}\right)+\left(1+\frac{1}{2.3}\right)+\left(1+\frac{1}{3.4}\right)+...+\left(1+\frac{1}{n\left(n+1\right)}\right)\)
\(a< n+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)\)
\(a< n+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(a< n+\left(1-\frac{1}{n+1}\right)< n+1\) (2)
Từ (1) và (2) suy ra phần nguyên của a là n
28 tháng 5 2017
Nguyễn Trần Thành ĐạtXuân Tuấn TrịnhHung nguyenHoang HungQuan Ace Legona giúp với
NT
25 tháng 12 2017
b.
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+....+\dfrac{2}{\left(n-1\right).n.\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)
TC
3 tháng 12 2017
Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)
\(\Rightarrow A< \dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{n^2-1}\)
\(\Rightarrow2A< \dfrac{2}{1.3}+\dfrac{2}{2.4}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(n-1\right)\left(n+1\right)}\)
\(\Rightarrow2A< 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n+1}\)
\(\Rightarrow2A< 1\)
\(\Rightarrow A< \dfrac{1}{2}< \dfrac{2}{3}\)
21 tháng 2 2018
a) \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Leftrightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\left(đúng\forall a;b\right)\)
Vậy bdt đã được cm
b) \(K=n\left(n+1\right)\left(n+2\right)\left(n+3\right)=\left(n^2+3n\right)\left(n^2+3n+2\right)\)
\(=\left(n^2+3n\right)^2+2\left(n^2+3n\right)\)
Ta có :
\(\left(n^2+3n\right)^2< \left(n^2+3n\right)^2+2\left(n^2+3n\right)< \left(n^2+3n\right)^2+2\left(n^2+3n\right)+1\)
\(\Leftrightarrow\left(n^2+3n\right)^2< \left(n^2+3n\right)^2+2\left(n^2+3n\right)< \left(n^2+3n+1\right)^2\)
Mà \(n^2+3n;n^2+3n+1\) là 2 số tn liên tiếp
\(\Rightarrow K\) không phải số chính phương
22 tháng 11 2018
a) Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
Ta có:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n}\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n}+1\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 2-\dfrac{1}{n}\)
\(\Rightarrow\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)< \dfrac{1}{2^2}\left(2-\dfrac{1}{2}\right)\)
\(\Rightarrow A< \dfrac{1}{2^2}.2-\dfrac{1}{2^2}.\dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{2^3}< \dfrac{1}{2}\)
Vậy \(A< \dfrac{1}{2}\left(Đpcm\right)\)
b) Đặt \(B=\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{\left(2n+1\right)^2}\)
Ta có:
\(B< \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(B< \dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(\dfrac{2n+1}{2n+1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}.\dfrac{2n}{2n+1}\)
\(B< \dfrac{2n}{4n+2}\)
\(B< \dfrac{2n}{2\left(2n+1\right)}\)
\(B< \dfrac{n}{2n+1}\)