Standard High contrast

Sequences & Series

Question 7 of 8

Geometric Progression, Series & Sums

Question

Find the sum of the geometric series \(9,\ 3,\ 1,\ \ldots,\ \dfrac{1}{6561}\)

Show solution Hide solution Fully worked — 12 steps
  1. Write down key terms
    \[ a_1 = 9;\quad a_2 = 3;\quad a_3 = 1;\quad a_n = \frac{1}{6561} \]
  2. Find \(r\) using \(r = \frac{a_2}{a_1}\)
    \[ r = \frac{a_2}{a_1} = \frac{3}{9} = \frac{1}{3}; \]
  3. Substitute the values of \(a_1\), \(a_n\) and \(r\) into \(a_n = a_1 r^{n-1}\) to find \(n\)
    \[ \frac{1}{6561} = (9)\left(\frac{1}{3}\right)^{n-1} \]
  4. Divide both sides by 9
    \[ \frac{1}{59049} = \left(\frac{1}{3}\right)^{n-1} \]
  5. Change \(\frac{1}{59049}\) to its exponential form whose base \(= r\)
    \[ \left(\frac{1}{3}\right)^{10} = \left(\frac{1}{3}\right)^{n-1} \]
  6. Equate the indices since they both have the same base
    \[ 10 = n - 1 \]
  7. Add 1 to both sides
    \[ 10 + 1 = n \]
  8. Add 10 and 1
    \[ 11 = n \]
  9. Substitute the values of \(a_1\), \(r\) and \(n\) into \(S_n = \frac{a_1(1-r^n)}{1-r}\)
    \[ S_{11} = \frac{9\left(1-\left(\frac{1}{3}\right)^{11}\right)}{1-\left(\frac{1}{3}\right)} \]
  10. Evaluate \(\left(\frac{1}{3}\right)^{11}\) then subtract \(\frac{1}{3}\) from \(1\)
    \[ = \frac{9\left(1-\frac{1}{177147}\right)}{\frac{2}{3}} \]
  11. Evaluate \(9\left(1-\frac{1}{177147}\right)\)
    \[ = \frac{\frac{177146}{19683}}{\frac{2}{3}} \]
  12. Divide \(\frac{177146}{19683}\) by \(\frac{2}{3}\)
    \[ S_{11} = \frac{88573}{6561} \]

    That is, the sum of the series is eighty-eight thousand, five hundred and seventy-three over six thousand, five hundred and sixty-one.