giải bàib= 1 phàn 3 mũ 2 1 phần 3 mũ 3

Lời giải:

A = \(\frac{1}{3} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ...... + \frac{1}{{{3^{20}}}}\)

3A = \[1 + \frac{1}{3} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ...... + \frac{1}{{{3^{19}}}}\]

Suy ra 3A – A = 1 −\(\frac{1}{{{3^{20}}}}\)

2A = \(\frac{{{3^{20}} - 1}}{{{3^{20}}}}\)

A = \(\frac{{{3^{20}} - 1}}{{{{2.3}^{20}}}}\)