Đặt P= 1/4^2+1/6^2+1/8^2+...1/2n^2

= > P= 1/2.(2/2.4+2/4.6+2/6.8+...+ 2/(2n-2).2n)

=> P= 1/2.(1/2-1/2n)

=> P= 1/2.1/2-1/2.1/2n

=> P = (1/4 -1/2.1/2n)(1/4

Vậy P<1/4 ( đcpcm)

1/4^2+1/6^2+...+1/(2n)^2<1/4

=>1/2^2+1/3^2+...+1/n^2<1

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{n\left(n-1\right)}=\dfrac{1}{2}-\dfrac{1}{n-1}< 1\)

=>ĐPCM

\(S=\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)

=>\(S< =\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)

=>\(S< =\dfrac{1}{4}\cdot\left(1-\dfrac{1}{n}\right)=\dfrac{1}{4}\cdot\dfrac{n-1}{n}< =\dfrac{1}{4}\)

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: 

\(a,P=\dfrac{1}{\left(2+1\right)\left(2+1-1\right):2}+\dfrac{1}{\left(3+1\right)\left(3+1-1\right):2}+...+\dfrac{1}{\left(2017+1\right)\left(2017+1-1\right):2}\\ P=\dfrac{1}{2\cdot3:2}+\dfrac{1}{3\cdot4:2}+...+\dfrac{1}{2017\cdot2018:2}\\ P=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)=2\cdot\dfrac{504}{1009}=\dfrac{1008}{1009}\)

\(b,\) Ta có \(\dfrac{1}{4^2}< \dfrac{1}{2\cdot4};\dfrac{1}{6^2}< \dfrac{1}{4\cdot6};...;\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{\left(2n-2\right)2n}\)

\(\Leftrightarrow VT< \dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2n-2\right)2n}\\ \Leftrightarrow VT< \dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{\left(2n-2\right)2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{2n}\right)< \dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\)

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)

@Luân Đào

\(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2n}\right)=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2n-1}+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2n}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n-1}+\frac{1}{2n}-\frac{1}{1}-\frac{1}{2}-....-\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\left(\text{đpcm}\right)\)

a) \(\dfrac{32}{\left(-2\right)^n}=4\)

\(\Rightarrow\left(-2\right)^n=8=\left(-2\right)^3\)

=> n = 3

b) \(\dfrac{8}{2^n}=2\)

\(\Rightarrow2^n=4=2^2\)

=> n = 2

c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)

=> 2n - 1 = 3

=> 2n = 4

=> n = 2

Giải:

a) \(\dfrac{32}{\left(-2\right)^n}=4\) 

\(\Rightarrow\left(-2\right)^n=32:4=8\) 

\(\Rightarrow\left(-2\right)^n=8\) 

Vì \(\left(-2\right)^n=2^3\) là ko thể nên n ∈ ∅

b) \(\dfrac{8}{2^n}=2\)

\(\Rightarrow2^n=8:2=4\) 

\(\Rightarrow2^n=4\) 

\(\Rightarrow2^n=2^2\) 

\(\Rightarrow n=2\) 

c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\) 

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\) 

\(\Rightarrow2n-1=3\rightarrow n=2\)