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Sequences & Series

Question 6 of 8

Geometric Progression, Series & Sums

Question

Find the sum of the geometric series \(2,\ -4,\ 8,\ \ldots,\ -4096\)

Show solution Hide solution Fully worked — 12 steps
  1. Write down key terms
    \[ a_1 = 2;\quad a_2 = -4;\quad a_3 = 8;\quad a_n = -4096 \]
  2. Find \(r\) using \(r = \frac{a_2}{a_1}\)
    \[ r = \frac{a_2}{a_1} = \frac{-4}{2} = -2; \]
  3. Substitute the values of \(a_1\), \(a_n\) and \(r\) into \(a_n = a_1 r^{n-1}\) to find \(n\)
    \[ -4096 = (2)(-2)^{n-1} \]
  4. Divide both sides by 2
    \[ -2048 = (-2)^{n-1} \]
  5. Change \(-2048\) to its exponential form whose base \(= r\)
    \[ (-2)^{11} = (-2)^{n-1} \]
  6. Equate the indices since they both have the same base
    \[ 11 = n - 1 \]
  7. Add 1 to both sides
    \[ 11 + 1 = n \]
  8. Add 11 and 1
    \[ 12 = n \]
  9. Substitute the values of \(a_1\), \(r\) and \(n\) into \(S_n = \frac{a_1(1-r^n)}{1-r}\)
    \[ S_{12} = \frac{2\left(1-(-2)^{12}\right)}{1-(-2)} \]
  10. Evaluate \((-2)^{12}\) then subtract \(-2\) from \(1\)
    \[ = \frac{2(1-4096)}{3} \]
  11. Evaluate \(2(1-4096)\)
    \[ = \frac{-8190}{3} \]
  12. Divide \(-8190\) by \(3\)
    \[ S_{12} = -2730 \]

    That is, the sum of the series is minus two thousand, seven hundred and thirty.