Giang Nguyễn
New member
Tính tổng riêng thứ n của chuỗi \(\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{{9^{n - 1}}}}} \)
A. \({s_n} = \frac{9}{8}(1 - \frac{1}{{{9^{n + 1}}}})\)
B. \({s_n} = \frac{1}{8}(1 - \frac{1}{{{9^{n }}}})\)
C. \({s_n} = (1 - \frac{1}{{{9^{n }}}})\)
D. \({s_n} = \frac{9}{8}(1 - \frac{1}{{{9^{n}}}})\)
A. \({s_n} = \frac{9}{8}(1 - \frac{1}{{{9^{n + 1}}}})\)
B. \({s_n} = \frac{1}{8}(1 - \frac{1}{{{9^{n }}}})\)
C. \({s_n} = (1 - \frac{1}{{{9^{n }}}})\)
D. \({s_n} = \frac{9}{8}(1 - \frac{1}{{{9^{n}}}})\)