\(1,\)

\(a,\) Với \(n=1\Leftrightarrow5+2\cdot1+1=8⋮8\left(đúng\right)\)

Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow5^k+2\cdot3^{k-1}+1⋮8\)

Với \(n=k+1\)

\(5^n+2\cdot3^{n-1}+1=5^{k+1}+2\cdot3^k+1\\ =5^k\cdot5+2\cdot3^k+1\\ =5^k\cdot2+2\cdot3^k+5^k\cdot3+1\\ =2\left(5^k+3^k\right)+5^k+2\cdot5^{k-1}+1+2\cdot3^{k-1}-2\cdot3^{k-1}\\ =2\left(5^k+3^k\right)+\left(5^k+2\cdot3^{k-1}+1\right)-2\left(3^{k-1}+5^{k-1}\right)\)

Vì \(5^k+3^k⋮\left(5+3\right)=8;5^{k-1}+3^{k-1}⋮\left(5+3\right)=8;5^k+2\cdot3^{k-1}+1⋮8\) nên \(5^{k+1}+2\cdot3^k+1⋮8\)

Theo pp quy nạp ta được đpcm

\(b,\) Với \(n=1\Leftrightarrow3^3+4^3=91⋮13\left(đúng\right)\)

Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow3^{k+2}+4^{2k+1}⋮13\)

Với \(n=k+1\)

\(3^{n+2}+4^{2n+1}=3^{k+3}+4^{2k+3}\\ =3^{k+2}\cdot3+16\cdot4^{2k+1}\\ =3^{k+2}\cdot3+3\cdot4^{2k+1}+13\cdot4^{2k+1}\\ =3\left(3^{k+2}+4^{2k+1}\right)+13\cdot4^{2k+1}\)

Vì \(3^{k+2}+4^{2k+1}⋮13;13\cdot4^{2k+1}⋮13\) nên \(3^{k+3}+4^{2k+3}⋮13\)

Theo pp quy nạp ta được đpcm

\(1,\)

\(c,C=6^{2n}+3^{n+2}+3^n\\ C=36^n+3^n\cdot9+3^n\\ C=\left(36^n-3^n\right)+\left(3^n\cdot9+2\cdot3^n\right)\\ C=\left(36^n-3^n\right)+3^n\cdot11\)

Vì \(36^n-3^n⋮\left(36-3\right)=33⋮11;3^n\cdot11⋮11\) nên \(C⋮11\)

\(d,D=1^n+2^n+5^n+8^n\)

Vì \(1^n+2^n+5^n⋮\left(1+2+5\right)=8;8^n⋮8\) nên \(D⋮8\)

\(2,\)

Ta thấy:\(1+2+...+2002=\left(2002+1\right)\left(2002-1+1\right):2=2003\cdot2002:2⋮11\left(2002⋮11\right)\)

Do đó \(1^{2002}+2^{2002}+...+2002^{2002}⋮1+2+...+2002⋮11\)

làm câu

Ta có:

\(A=(1-\frac{1}{1+2})(1-\frac{1}{1+2+3})(1-\frac{1}{1+2+3+4}) ...(1-\frac{1}{1+2+3+...+n}) \)

Xét công thức tổng quát ta có:

\(1-\frac{1}{1+2+3+...+n}=\frac{2+3+...n.}{1+2+3+..+n} =\frac{n(n+1)-2}{2}:\frac{n(n+1)}{2}=\frac{(n+2)(n-1)}{n(n+1)} \)

Áp dụng ct tổng quá ta có:

A=\(\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{(n-1)(n+2)}{n(n+1)} \)=\(\frac{(1.2.3...(n-1))(4.5.6...(n+2))}{(2.3.4...n)(3.4.5...(n+1))} \)=\(\frac{n+2}{3n} \)

=>A:B=\(\frac{n+2}{3n}:\frac{n+2}{n}=\frac{1}{3} \)

Bài 2  : 

a)    C = ( n + 1 )( n + 2 )( n + 3 )( n + 4 )

<=> C = [( n + 1 ).( n + 4 )].[( n + 2 ).( n + 3 )] + 1

<=> C = ( n² + 5n + 4 ).( n² + 5n + 6 ) + 1 

Đặt t = n² + 5n + 5

Suy ra : C = ( t - 1 ).( t + 1 ) + 1

         => C = t² - 1 + 1

       <=> C = t²    hay C = ( n² + 5n + 5 )²

Vì n thuộc Z => n² + 5n + 5 thuộc Z => C là số chính phương 

                                                                             ( đpcm )

b)     E = n² + ( n + 1 )² + n² ( n + 1 )²

 <=> E = n² - 2n( n + 1 ) + ( n + 1 )² + 2n( n + 1 ) + n²( n +1 )²

 <=> E = [ n - ( n + 1 )]² + 2n( n + 1 ) + [ n( n + 1 )]²

 <=> E = ( n - n - 1 )² + 2n( n + 1 ) + [ n( n + 1 )]²

 <=> E = 1² + 2.1.n( n + 1 ) + [ n( n + 1 )]²

 <=> E = [ n( n + 1 ) + 1 ]²

 <=> E = ( n² + n + 1 )²

Vì n thuộc Z => n² + n + 1 thuộc Z => E là số chính phương

                                                                        ( đpcm )

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n+1\right)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}\)

b) \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)

         \(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

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

Bài 1:

\(\dfrac{5}{x} - \dfrac{y}{3} =\dfrac{1}{6}\)

\(\Rightarrow\dfrac{1}{6}+\dfrac{y}{3}=\dfrac{5}{x}\)

\(\Rightarrow\dfrac{1}{6}+\dfrac{2y}{6}=\dfrac{5}{x}\)

\(\Rightarrow1+\dfrac{2y}{6}=\dfrac{5}{x}\)

\(\Rightarrow x.\left(1+2y\right)=30\)

Vì \(2y\) chẵn nên \(1+2y\) lẻ

\(\Rightarrow1+2y\in\left\{\pm1;\pm3;\pm5;\pm30\right\}\)

\(\Rightarrow x\in\left\{\pm10;\pm30;\pm6;\pm2\right\}\)

Bài 2:

\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{\left(2n-2\right).2n}\)

\(=\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{\left(2n-2\right).2n}\right).\dfrac{1}{2}\)

\(=\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{12}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right).\dfrac{1}{2}\)

\(=\left(\dfrac{1}{2}-\dfrac{1}{2n}\right).\dfrac{1}{2}\)

\(=\dfrac{1}{4}-\dfrac{1}{2n.2}< \dfrac{1}{4}\)

\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(đpcm\right)\)

Câu B dấu chấm than là kí hiệu gì thế bạn?