JAMB 1992 · UME · Q18

Find the sum to infinity of the series 3+2+43+89+1627+…3 + 2 + \frac43 + \frac89 + \frac{16}{27} + \dots

Worked solution (try it first)
  1. This is a G.P. with a=3a = 3 and r=2÷3=23r = 2 \div 3 = \frac23.
  2. Since rr is between −1-1 and 1, the sum to infinity is S∞=a1−rS_\infty = \dfrac{a}{1 - r}.
  3. So S∞=3÷13=9S_\infty = 3 \div \frac13 = 9, option D.

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