Với mọi số tự nhiên \(n \ge 1 \), tổng \({S_n} = \frac{1}{3} + \frac{2}{{{3^2}}} + ... + \frac{n}{{{3^n}}} \) là:
A.\({S_n} = \frac{{2n + 3}}{{{{4.3}^n}}}\)
B.\({S_n} = \frac{3}{4} + \frac{{2n + 3}}{{{{4.3}^n}}}\)
C.\({S_n} = \frac{{2n + 3}}{{{{4.3}^n}}} - \frac{3}{4}\)
D.\({S_n} = \frac{3}{4} - \frac{{2n + 3}}{{{{4.3}^n}}}\)
A.\({S_n} = \frac{{2n + 3}}{{{{4.3}^n}}}\)
B.\({S_n} = \frac{3}{4} + \frac{{2n + 3}}{{{{4.3}^n}}}\)
C.\({S_n} = \frac{{2n + 3}}{{{{4.3}^n}}} - \frac{3}{4}\)
D.\({S_n} = \frac{3}{4} - \frac{{2n + 3}}{{{{4.3}^n}}}\)
Loga
Hóa Học lớp 12
