\(C=3^{n+2}-2^{n+2}+3^n-2^n\)

\(C=\left(3^{n+2}-2^{n+2}\right)+\left(3^n-2^n\right)\)

\(\Rightarrow C=1^{n+2}+1^n\) (với n \(\in\)N*)

Ta có công thức Cơ số có tận cùng bằng 1 thì mũ lên bao nhiêu cũng bằng 1.(với n \(\in\)N*)

Vì  n \(\in\)N* \(\Rightarrow C=1^{n+2}+1^n=\left(...1\right)+\left(...1\right)=\left(...2\right)\)

Hơi lâu nên đợi anh chút

\(D=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{3n+1}{3^n}\)

\(\Rightarrow3D=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{3n+1}{3^{n-1}}\)

\(\Rightarrow3D-D=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{3n+1}{3^{n-1}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{3n+1}{3^n}\right)\)

\(\Rightarrow2D=4+1+\frac{1}{3}+...+\frac{1}{3^{n-2}}-\frac{3n+1}{3^n}\)

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

\(\Rightarrow3M=12+3+1+...+\frac{1}{3^{n-3}}\)

\(\Rightarrow3M-M=\left(12+3+1+...+\frac{1}{3^{n-3}}\right)-\left(4+1+\frac{1}{3}+...+\frac{1}{3^{n-2}}\right)\)

\(\Rightarrow2M=11-\frac{1}{3^{n-2}}< 11\)

\(\Rightarrow2M< 11\)

\(\Rightarrow M< \frac{11}{2}\)

\(\Rightarrow2D< \frac{11}{2}\)

\(\Rightarrow D< \frac{11}{4}\left(đpcm\right)\)

1.

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

2.

\(\lim \frac{2n^3-4n^2+3n+7}{n^3-7n+5}=\lim \frac{\frac{2n^3-4n^2+3n+7}{n^3}}{\frac{n^3-7n+5}{n^3}}=\lim \frac{2-\frac{4}{n}+\frac{3}{n^2}+\frac{7}{n^3}}{1-\frac{7}{n^2}+\frac{5}{n^3}}=\frac{2}{1}=2\)

3.

\(\lim (\frac{2n^3}{2n^2+3}+\frac{1-5n^2}{5n+1})=\lim (n-\frac{3n}{2n^2+3}+\frac{1}{5}-n-\frac{1}{5n+1})\)

\(=\frac{1}{5}-\lim (\frac{3n}{2n^2+3}+\frac{1}{5n+1})=\frac{1}{5}-\lim (\frac{3}{2n+\frac{3}{n}}+\frac{1}{5n+1})=\frac{1}{5}-0=\frac{1}{5}\)

4.

\(\lim \frac{1+3^n}{4+3^n}=\lim (1-\frac{3}{4+3^n})=1-\lim \frac{3}{4+3^n}=1-0=1\)

5.

\(\lim \frac{4.3^n+7^{n+1}}{2.5^n+7^n}=\lim \frac{\frac{4.3^n+7^{n+1}}{7^n}}{\frac{2.5^n+7^n}{7^n}}\)

\(=\lim \frac{4.(\frac{3}{7})^n+7}{2.(\frac{5}{7})^n+1}=\frac{7}{1}=7\)

a)\(\left(\frac{1}{5}\right)^{3n-1}=\frac{1}{25}\)

\(\Leftrightarrow\left(\frac{1}{5}\right)^{3n-1}=\left(\frac{1}{5}\right)^2\)

\(\Leftrightarrow3n-1=2\)

\(\Leftrightarrow3n=3\)

\(\Leftrightarrow n=1\)

b)\(\left(\frac{4}{7}\right)^{n+2}=\frac{7}{4}\)

\(\Leftrightarrow\left(\frac{4}{7}\right)^{n+2}=\left(\frac{4}{7}\right)^{-1}\)

\(\Leftrightarrow n+2=-1\)

\(\Leftrightarrow n=-3\)

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

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

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

\(\Leftrightarrow-n+1=-3\)

\(\Leftrightarrow n=-4\)

c)\(\left(0,7\right)^{3n+1}=10^3:7^3\)

\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{10}{7}\right)^3\)

\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{7}{10}\right)^{-3}\)

\(\Leftrightarrow3n+1=-3\)

\(\Leftrightarrow3n=-4\)

\(\Leftrightarrow n=-\frac{4}{3}\)

a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)

\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)

b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)

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

\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)

\(=\frac{5}{4}.\frac{4n}{12n+9}\)

\(=\frac{5n}{12n+9}\)

( sai đề )

      2. Goi d la uoc chung lon nhat cua n va n+1 thi \(n⋮d\) va \(n+1⋮d\)

        \(\Rightarrow n+1-n⋮d\Rightarrow1⋮d\Rightarrow d\in\left[1;-1\right]\)

        Vay \(\frac{n}{n+1}\)la phan so toi gian