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\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2013^2}\)
Ta có ;
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
...
\(\dfrac{1}{2013^2}< \dfrac{1}{2012.2013}\)
\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2012.2013}\)
\(\Rightarrow B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2013}\)
\(\Leftrightarrow B< 1-\dfrac{1}{2013}\)
\(\Rightarrow B< \dfrac{2012}{2013}\)
Lại có : \(\dfrac{2012}{2013}< \dfrac{3}{4}\)
\(\Rightarrow B< \dfrac{3}{4}\)
* Chắc vậy, sai thì thôg cảm ^^ *
Còn j k hiểu thì ib nha
Ta có: \(\dfrac{1}{2^2}>\dfrac{1}{2\cdot3}\)
\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}\)
\(\dfrac{1}{4^2}>\dfrac{1}{4\cdot5}\)
..................
\(\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}\)
\(\Rightarrow\) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{9\cdot10}\)
\(\Rightarrow\) \(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow\) \(A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow\) \(A>\dfrac{5}{10}-\dfrac{1}{10}\)
\(\Rightarrow\) \(A>\dfrac{2}{5}\) (1)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3\cdot4}\)
...................
\(\dfrac{1}{9^2}< \dfrac{1}{8\cdot9}\)
\(\Rightarrow\) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}\)
\(\Rightarrow\) \(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow\) \(A< 1-\dfrac{1}{9}\)
\(\Rightarrow\) \(A< \dfrac{9}{9}-\dfrac{1}{9}\)
\(\Rightarrow\) \(A< \dfrac{8}{9}\) (2)
Từ (1) và (2) ta được: \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
Vậy \(\dfrac{2}{5}< A< \dfrac{8}{9}\).
Mà đề phần kết luận sai nhé, nếu \(\dfrac{1}{n^2}\) thì A đâu lớn hơn \(\dfrac{2}{5}\), phải thay \(\dfrac{1}{n^2}\) thành \(\dfrac{1}{9^2}\) nha
\(\dfrac{1}{1\cdot2}>\dfrac{1}{2^2}>\dfrac{1}{2\cdot3},\dfrac{1}{2\cdot3}>\dfrac{1}{3^2}>\dfrac{1}{3\cdot4},...,\dfrac{1}{8\cdot9}>\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}>\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\) \(\Rightarrow1-\dfrac{1}{9}>A>\dfrac{1}{2}-\dfrac{1}{10}\) \(\Rightarrow\dfrac{8}{9}>A>\dfrac{2}{5}\)
Ta có \(M=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+59}\)
= \(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{59\cdot60}\)
= \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\)
= \(\dfrac{1}{3}-\dfrac{1}{60}=\dfrac{19}{60}< \dfrac{40}{60}=\dfrac{2}{3}\)
Vậy M < \(\dfrac{2}{3}\)
Lời giải:
Ta có:
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{(2n)^2}< \frac{1}{4^2-1}+\frac{1}{6^2-1}+\frac{1}{8^2-1}+...+\frac{1}{(2n)^2-1}(*)\)
Mà:
\(\frac{1}{4^2-1}+\frac{1}{6^2-1}+\frac{1}{8^2-1}+...+\frac{1}{(2n)^2-1}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n-1)(2n+1)}\)
\(=\frac{1}{2}\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2n+1}\right)\)
\(< \frac{1}{6}< \frac{1}{4}(**)\)
Từ \((*);(**)\Rightarrow N< \frac{1}{4}\) (đpcm)
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
ta có : \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
............
\(\dfrac{1}{n^2}=\dfrac{1}{n.n}< \dfrac{1}{n.(n-1)}\)
đặt tổng đó là A
A=\(\dfrac{1}{2^n}+\dfrac{1}{2^n}+.....+\dfrac{1}{2^n}\)
=\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-....-\dfrac{1}{n-1}+\dfrac{1}{n}\)
=\(\dfrac{1-1}{n}\)
=\(\dfrac{n-1}{n}\)<1
vậy A lớn hơn 1
nhung vay A<1 cho sao cau A>1
chac cau lon nhi
ma o tren cau dung day
cam on cau rat nhieu!