[tex]u_{n}=u_{1}+(n-1)3=3n-2[/tex]
=> [tex]S_{n}=-2\left ( \frac{1}{3}+\frac{1}{3^2} +...+\frac{1}{3^n}\right )+3\left ( \frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{n}{3^n} \right )[/tex]
[tex]A=\left ( \frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{n}{3^n} \right ) \Rightarrow 3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{n}{3^{n-1}}[/tex]
=> [tex]3A-A=2A=1+\left ( \frac{2}{3}-\frac{1}{3} \right )+\left ( \frac{3}{3^2}-\frac{2}{3^2} \right )+...+\left ( \frac{n}{3^{n-1}}-\frac{n-1}{3^{n-1}} \right )-\frac{n}{3^n}[/tex]
=> [tex]2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-1}}-\frac{n}{3^n}=\frac{1-\left( \frac{1}{3} \right )^n}{1-\frac{1}{3}}-\frac{n}{3^n}[/tex]
=> [tex]S_{n}=\frac{-2}{3}.\frac{1-\left(\frac{1}{3} \right )^n}{1-\frac{1}{3}}+\frac{3}{2}.\left (\frac{1-\left( \frac{1}{3} \right )^n}{1-\frac{1}{3}}-\frac{n}{3^n} \right )=\frac{5}{6}.\frac{1-\left(\frac{1}{3} \right )^n}{1-\frac{1}{3}}-\frac{1}{2}.\frac{n}{3^n}[/tex]
=> [tex]\lim_{x \rightarrow \infty }{\frac{5}{6}.\frac{1-\left(\frac{1}{3} \right )^n}{1-\frac{1}{3}}-\frac{1}{2}.\frac{n}{3^n}}=\lim_{x \rightarrow \infty }{\frac{5.3^n-5-2n}{4.3^n}}=\frac{5}{4}[/tex]
Có gì sai xót em tự coi mà fix lại nha chứ hướng đi oke rồi đó
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
