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

\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)

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

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

\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)

a) \(1+2+3+4+...+n\)

\(=\left(n+1\right)\left[\left(n-1\right):1+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right):2\)

\(=n\left(n+1\right):2\)

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

b) \(2+4+6+..+2n\)

\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)

\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right)\)

\(=n\left(n+1\right)\)

c) \(1+3+5+...+\left(2n+1\right)\)

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

\(=\left(2n+1+1\right)\left[\left(2n-1-1\right):2+1\right]:2\)

\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)

\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right)\)

\(=n\left(n+1\right)\)

d) \(1+4+7+10+...+2005\)

\(=\left(2005+1\right)\left[\left(2005-1\right):3+1\right]:2\)

\(=2006\cdot\left(2004:3+1\right):2\)

\(=2006\cdot\left(668+1\right):2\)

\(=1003\cdot669\)

\(=671007\)

e) \(2+5+8+...+2006\)

\(=\left(2006+2\right)\left[\left(2006-2\right):3+1\right]:2\)

\(=2008\cdot\left(2004:3+1\right):2\)

\(=1004\cdot\left(668+1\right)\)

\(=1004\cdot669\)

\(=671676\)

g) \(1+5+9+...+2001\)

\(=\left(2001+1\right)\left[\left(2001-1\right):4+1\right]:2\)

\(=2002\cdot\left(2000:4+1\right):2\)

\(=1001\cdot\left(500+1\right)\)

\(=1001\cdot501\)

\(=501501\)

a: \(\Leftrightarrow2n^2+n-2n-1+3⋮2n+1\)

\(\Leftrightarrow2n+1\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{0;-1;1;-2\right\}\)

b: \(\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\)

\(\Leftrightarrow n-2\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{3;1;5;-1\right\}\)

c: \(\Leftrightarrow10n^2-15n+8n-12+7⋮2n-3\)

\(\Leftrightarrow2n-3\in\left\{1;-1;7;-7\right\}\)

hay \(n\in\left\{2;1;5;-2\right\}\)

d: \(\Leftrightarrow2n^2-n+4n-2+5⋮2n-1\)

\(\Leftrightarrow2n-1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{1;0;3;-2\right\}\)

2.a)n^5+1⋮n^3+1

⇒n^2.(n^3+1)-n^2+1⋮n^3+1

⇒1⋮n^3+1

⇒n^3+1ϵƯ(1)={1}

ta có :n^3+1=1

n^3=0

n=0

Vậy n=0

b)n^5+1⋮n^3+1

Vẫn làm y như bài trên nhưng vì nϵZ⇒n=0

Bữa sau giải bài 3 mình buồn ngủ quá!!!!!!!!

\(1,\)

\(a,\) 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 \(1^{2002}+2^{2002}+...+2002^{2002}⋮1+2+...+2002⋮11\)