Tính tổng:
a, S = \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)+ \(\frac{1}{256}\)
b, A = 1 + \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{10}\)+ \(\frac{1}{15}\)+ \(\frac{1}{21}\)+ \(\frac{1}{28}\)+ \(\frac{1}{36}\)+ \(\frac{1}{45}\)