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1: \(\lim_{}\frac{1+2+\cdots+n}{n^2+2n+11}\)

\(=\lim_{}\frac{\frac{n\left(n+1\right)}{2}}{n^2+2n+11}=\lim_{}\frac{n\left(n+1\right)}{2\left(n^2+2n+11\right)}\)

\(=\lim_{}\frac{n^2+n}{2n^2+4n+22}=\lim_{}\frac{1+\frac{1}{n}}{2+\frac{4}{n}+\frac{22}{n^2}}=\frac12\)

2: \(\lim_{}\frac{n\cdot\sqrt{1+2+3+\cdots+n}}{n^2+n+1}\)

\(=\lim_{}\frac{n\cdot\sqrt{\frac{n\left(n+1\right)}{2}}}{n^2+n+1}=\lim_{}\frac{n\cdot\sqrt{n^2+n}}{\sqrt2\cdot\left(n^2+n+1\right)}\)

\(=\lim_{}\frac{n^2\left(\sqrt{1+\frac{1}{n}}\right)}{n^2\cdot\sqrt2\cdot\left(1+\frac{1}{n}+\frac{1}{n^2}\right)}=\lim_{}\frac{\sqrt{1+\frac{1}{n}}}{\sqrt2\left(1+\frac{1}{n}+\frac{1}{n^2}\right)}=\frac{1}{\sqrt2}=\frac{\sqrt2}{2}\)

3: \(\lim_{}\frac{1^2+2^2+\cdots+n^2}{n^3+3n+2}\)

\(=\lim_{}\frac{\frac{n\left(n+1\right)\left(2n+1\right)}{6}}{n^3+3n+2}=\lim_{}\frac{n\left(n+1\right)\left(2n+1\right)}{6\left(n^3+3n+2\right)}\)

\(=\lim_{}\frac{1\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6\left(1+\frac{3}{n^2}+\frac{2}{n^3}\right)}=\frac{1\cdot1\cdot2}{6\cdot1}=\frac26=\frac13\)

4: \(\lim_{}\frac{1^2+2^3+3^3+\cdots+n^3}{2n^4-3n^3+8}\)

\(=\lim_{}\frac{\left(1+2+3+\cdots+n\right)^2}{2n^4-3n^3+8}=\lim_{}\frac{\left\lbrack\frac{n\left(n+1\right)}{2}\right\rbrack^2}{n^4\left(2-\frac{3}{n}+\frac{8}{n^4}\right)}\)

\(=\lim_{}\frac{n^2\cdot\left(n+1\right)^2}{4n^4\left(2-\frac{3}{n}+\frac{8}{n^4}\right)}=\lim_{}\frac{n^2\cdot n^2\cdot\left\lbrack1\cdot\left(1+\frac{1}{n}\right)^2\right.]^{}}{4n^4\left(2-\frac{3}{n}+\frac{8}{n^4}\right)}\)

\(=\lim_{}\frac{\left(1+\frac{1}{n}\right)^2}{4\left(2-\frac{3}{n}+\frac{8}{n^4}\right)}=\frac{\left(1+0\right)^2}{4\left(2-0+0\right)}=\frac{1}{4\cdot2}=\frac18\)

5: \(1\cdot2+2\cdot3+\cdots+n\left(n+1\right)\)

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

\(=\left(1^2+2^2+\cdots+n^2\right)+\left(1+2+\cdots+n\right)\)

\(=\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}=\frac{n\left(n+1\right)\left(2n+1\right)+3n\left(n+1\right)}{6}=\frac{n\left(n+1\right)\left(2n+4\right)}{6}=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

\(\lim_{}\frac{1\cdot2+2\cdot3+\cdots+n\left(n+1\right)}{4n^3+2n^2-n+3}\)

\(=\lim_{}\frac{n\left(n+1\right)\left(n+2\right)}{3\left(4n^3+2n^2-n+3\right)}=\lim_{}\frac{1\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)}{3\left(4+\frac{2}{n}-\frac{1}{n^2}+\frac{3}{n^3}\right)}=\frac{1}{3\cdot4}=\frac{1}{12}\)

6: \(\lim_{}\frac{n\cdot\sqrt{2+4+6+\cdots+2n}}{3n^2+n-1}\)

\(=\lim_{}\frac{n\cdot\sqrt{2\left(1+2+\cdots+n\right)}}{3n^2+n-1}\)

\(=\lim_{}\frac{n\cdot\sqrt{\frac{2n\left(n+1\right)}{2}}}{3n^2+n-1}=\lim_{}\frac{n\cdot\sqrt{n^2+n}}{3n^2+n-1}\)

\(=\lim_{}\frac{n\cdot n\cdot\sqrt{1+\frac{1}{n}}}{n^2\left(3+\frac{1}{n}-\frac{1}{n^2}\right)}=\lim_{}\frac{\sqrt{1+\frac{1}{n}}}{3+\frac{1}{n}-\frac{1}{n^2}}=\frac13\)

7: \(1+3+\cdots+\left(2n-1\right)\)

Số số hạng của dãy số là:

(2n-1-1):2+1=(2n-2):1+1=n-1+1=n(số)

Tổng của dãy số là: \(\left(2n-1+1\right)\cdot\frac{n}{2}=2n\cdot\frac{n}{2}=n^2\)

\(\lim_{}\frac{n\cdot\sqrt{1+3+5+\cdots+2n-1}}{6n^2+n+1}\)

\(=\lim_{}\frac{n\cdot\sqrt{n^2}}{6n^2+n+1}=\lim_{}\frac{n^2}{6n^2+n+1}=\lim_{}\frac{1}{6+\frac{1}{n}+\frac{1}{n^2}}=\frac16\)

9: \(\lim_{}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{n\left(n+1\right)}\right)\)

\(=\lim_{}\left(1-\frac12+\frac12-\frac13+\cdots+\frac{1}{n}-\frac{1}{n+1}\right)=\lim_{}\left(1-\frac{1}{n+1}\right)\)

\(=1-0=1\)

10: \(\lim_{}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\lim_{}\left\lbrack\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\right\rbrack\)

\(=\lim_{}\left\lbrack\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\right\rbrack\)

\(=\lim_{}\left\lbrack\frac12\left(1-\frac{1}{2n+1}\right)\right\rbrack=\lim_{}\left\lbrack\frac12\cdot\frac{2n}{2n+1}\right\rbrack=\lim_{}\frac{n}{2n+1}=\lim_{}\frac{1}{2+\frac{1}{n}}=\frac12\)