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Sum of the First nn Terms of a Geometric Progression

Consider a finite geometric progression b1,b2,b3,…,bn−1,bnb_1, b_2, b_3, \ldots, b_{n-1}, b_n. Let SnS_n denote the sum of the terms:

Sn=b1+b2+b3+…+bn−1+bn.(∗)S_n = b_1 + b_2 + b_3 + \ldots + b_{n-1} + b_n. \qquad (*)

Derivation of the Formula

First consider a specific problem. Take the geometric progression 1,2,22,…,262,2631, 2, 2^2, \ldots, 2^{62}, 2^{63} and find the sum S64S_{64}:

S64=1+2+22+…+262+263.S_{64} = 1 + 2 + 2^2 + \ldots + 2^{62} + 2^{63}.

Multiply both sides by the common ratio 2:

2S64=2+22+…+262+263+264.2S_{64} = 2 + 2^2 + \ldots + 2^{62} + 2^{63} + 2^{64}.

Subtract S64S_{64} from 2S642S_{64}:

2S64−S64=−1+0+0+…+0+264.2S_{64} - S_{64} = -1 + 0 + 0 + \ldots + 0 + 2^{64}.

Therefore S64=264−1S_{64} = 2^{64} - 1.

ℹNote — The Legend of the Chessboard

This sequence is connected to an ancient legend. An Indian sage who invented chess asked the king for a seemingly modest reward: 1 grain of wheat for the first square of the chessboard, 2 for the second, 4 for the third, and so on — each subsequent square receiving twice as many grains as the previous one.

The total number of grains requested equals S64=264−1=18 446 744 073 709 551 615S_{64} = 2^{64} - 1 = 18\,446\,744\,073\,709\,551\,615.

Now derive the general formula. Rewrite (∗)(*) as:

Sn=b1+b1q+b1q2+b1q3+…+b1qn−2+b1qn−1.S_n = b_1 + b_1 q + b_1 q^2 + b_1 q^3 + \ldots + b_1 q^{n-2} + b_1 q^{n-1}.

Multiply both sides by qq:

Snq=b1q+b1q2+b1q3+b1q4+…+b1qn−1+b1qn.S_n q = b_1 q + b_1 q^2 + b_1 q^3 + b_1 q^4 + \ldots + b_1 q^{n-1} + b_1 q^n.

Subtract SnS_n from SnqS_n q:

Snq−Sn=b1qn−b1.S_n q - S_n = b_1 q^n - b_1.

So Sn(q−1)=b1(qn−1)S_n(q - 1) = b_1(q^n - 1).

⚡Theorem — Sum of the First n Terms of a Geometric Progression

For q≠1q \neq 1:

Sn=b1(qn−1)q−1S_n = \frac{b_1(q^n - 1)}{q - 1}

If q=1q = 1, all terms equal the first term, so Sn=nb1S_n = nb_1.

Example

✎Example — Finding the First Term and Common Ratio from the Sum Formula

Problem. For any natural number nn, the sum of the first nn terms of a geometric progression can be computed by the formula Sn=10(2n−1)S_n = 10(2^n - 1). Find the first term and common ratio.

Solution. Let b1b_1 be the first term and qq the common ratio. Then b1=S1=10(21−1)=10b_1 = S_1 = 10(2^1 - 1) = 10; b1+b2=S2=10(22−1)=30b_1 + b_2 = S_2 = 10(2^2 - 1) = 30. So b2=30−b1=20b_2 = 30 - b_1 = 20; q=b2b1=2010=2q = \dfrac{b_2}{b_1} = \dfrac{20}{10} = 2.

Answer: b1=10b_1 = 10, q=2q = 2.

Exercises

✏Exercise — Sum of Geometric Progression Problems

34.1. Find the sum of the first nn terms of the geometric progression (bn)(b_n) with common ratio qq if:

  1. b1=0.6b_1 = 0.6, q=2q = 2, n=5n = 5
  2. b1=−4b_1 = -4, q=−1q = -1, n=10n = 10
  3. b1=−9b_1 = -9, q=3q = \sqrt{3}, n=6n = 6

34.3. Find the sum of the first five terms of the geometric progression:

  1. 12,72,432,…12, 72, 432, \ldots
  2. 116,−18,14,…\dfrac{1}{16}, -\dfrac{1}{8}, \dfrac{1}{4}, \ldots

34.10. The sum of the first three terms of a geometric progression is 516 and the first term is 12. Find the common ratio.

34.11. The sum of a finite geometric progression is 8191. Find the number of terms if the first term b1=1b_1 = 1 and the common ratio q=2q = 2.