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:

\(B< \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)

\(B< \dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(B< \dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

\(B< \dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)

\(B< \dfrac{1}{2}\left(\dfrac{2n+1}{2n+1}-\dfrac{1}{2n+1}\right)\)

\(B< \dfrac{1}{2}.\dfrac{2n}{2n+1}\)

\(B< \dfrac{2n}{4n+2}\)

\(B< \dfrac{2n}{2\left(2n+1\right)}\)

\(B< \dfrac{n}{2n+1}\)

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)

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

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

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

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: 

\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)

\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)

\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)

\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)

\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)

\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)

Lời giải:

Ta có: \(4+(2n-1)^4=[(2n-1)^2+2]^2-[2(2n-1)]^2\)

\(=[(2n-1)^2+2-2(2n-1)][(2n-1)^2+2+2(2n-1)]\)

\(\Rightarrow \frac{2n-1}{4+(2n-1)^4}=\frac{2n-1}{[(2n-1)^2+2-2(2n-1)][(2n-1)^2+2+2(2n-1)]}\)

\(=\frac{1}{4}\left(\frac{1}{(2n-1)^2+2-2(2n-1)}-\frac{1}{(2n-1)^2+2+2(2n-1)}\right)\)

Do đó:

\(\frac{1}{4+1^4}=\frac{1}{4}(1-\frac{1}{5})\)

\(\frac{3}{4+3^4}=\frac{1}{4}(\frac{1}{5}-\frac{1}{17})\)

\(\frac{5}{4+5^4}=\frac{1}{4}(\frac{1}{17}-\frac{1}{37})\)

......

Do đó:

\(\frac{1}{4+1^4}+\frac{3}{4+3^4}+...+\frac{2n-1}{4+(2n-1)^4}=\frac{1}{4}(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{17}+...+\frac{1}{(2n-1)^2+2-2(2n-1)}-\frac{1}{(2n-1)^2+2+2(2n-1)})\)

\(=\frac{1}{4}(1-\frac{1}{(2n-1)^2+2+2(2n-1)})=\frac{1}{4}(1-\frac{1}{(2n-1+1)^2+1})\)

\(=\frac{1}{4}(1-\frac{1}{4n^2+1})=\frac{n^2}{4n^2+1}\)

Ta có đpcm.

n=1 ; \(\dfrac{1}{4+1^4}=\dfrac{1}{5}=\dfrac{1^2}{4.^2+1}=\dfrac{1}{5};dung\)

giả sử n =k đúng \(\Leftrightarrow S=\dfrac{1}{4+1^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}=\dfrac{k^2}{4k^2+1}\) (*)

cần c/m đúng n =k+1 ;

c/m

với n=k+1

\(S=\left(\dfrac{1}{4+1^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}\right)+\dfrac{2\left(k+1\right)-1}{4+\left(2\left(k+1\right)-1\right)^4}\)

từ (*) =>\(S=\dfrac{k^2}{4k^2+1}+\dfrac{2\left(k+1\right)-1}{4+\left(2\left(k+1\right)-1\right)^4}\)

\(k+1=t\Leftrightarrow k=t-1\)

\(S=\dfrac{t^2-2t+1}{4\left(t^2-2t+1\right)+1}+\dfrac{2t-1}{4+\left(2t-1\right)^4}\)

\(S=\dfrac{t^2-2t+2}{4t^2-8t+5}+\dfrac{2t-1}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}=\dfrac{\left(t^2-2t+1\right)\left(4t^2+1\right)+2t-1}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}\)\(S=\dfrac{t^2\left(4t^2-8t+5\right)}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}=\dfrac{t^2}{\left(4t^2+1\right)}=\dfrac{\left(k+1\right)^2}{4\left(k+1\right)^2+1}\)

Vậy tổng trên đúng với k +1

theo Quy nạp ta có dpcm

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

2.

\(4n^3+n+3=4n^3+2n^2+2n-2n^2-n-1+4=2n\left(2n^2+n+1\right)-\left(2n^2+n+1\right)+4\)-Để \(\left(4n^3+n+3\right)⋮\left(2n^2+n+1\right)\) thì \(4⋮\left(2n^2+n+1\right)\)

\(\Leftrightarrow2n^2+n+1\in\left\{1;-1;2;-2;4;-4\right\}\) (do n là số nguyên)

*\(2n^2+n+1=1\Leftrightarrow n\left(2n+1\right)=0\Leftrightarrow n=0\) (loại) hay \(n=\dfrac{-1}{2}\) (loại)

*\(2n^2+n+1=-1\Leftrightarrow2n^2+n+2=0\) (phương trình vô nghiệm)

\(2n^2+n+1=2\Leftrightarrow2n^2+n-1=0\Leftrightarrow n^2+n+n^2-1=0\Leftrightarrow n\left(n+1\right)+\left(n+1\right)\left(n-1\right)=0\Leftrightarrow\left(n+1\right)\left(2n-1\right)=0\)

\(\Leftrightarrow n=-1\) (loại) hay \(n=\dfrac{1}{2}\) (loại)

\(2n^2+n+1=-2\Leftrightarrow2n^2+n+3=0\) (phương trình vô nghiệm)

\(2n^2+n+1=4\Leftrightarrow2n^2+n-3=0\Leftrightarrow2n^2-2n+3n-3=0\Leftrightarrow2n\left(n-1\right)+3\left(n-1\right)=0\Leftrightarrow\left(n-1\right)\left(2n+3\right)=0\)\(\Leftrightarrow n=1\left(nhận\right)\) hay \(n=\dfrac{-3}{2}\left(loại\right)\)

-Vậy \(n=1\)

1. \(x^2+y^2=z^2\)

\(\Rightarrow x^2+y^2-z^2=0\)

\(\Rightarrow\left(x-z\right)\left(x+z\right)+y^2=0\)

-TH1: y lẻ \(\Rightarrow x-z;x+z\) đều lẻ.

\(x+3z-y=x+z-y+2x\) chia hết cho 2. \(\Rightarrow\)Hợp số.

-TH2: y chẵn \(\Rightarrow\)1 trong hai biểu thức \(x-z;x+z\) chia hết cho 2.

*Xét \(\left(x-z\right)⋮2\):

\(x+3z-y=x-z+4z-y\) chia hết cho 2. \(\Rightarrow\)Hợp số.

*Xét \(\left(x+z\right)⋮2\):

\(x+3z-y=x+z+2z-y\) chia hết cho 2 \(\Rightarrow\)Hợp số.