33
Sum of the first n terms of a geometric progression
Textbook: pp. 167–170
GoalDerive and use the formula for the sum of the first n terms of a geometric progression.
New words
sum of terms · hadlar yig‘indisiratio q ≠ 1 · maxraj q ≠ 1S_n formula · S_n formulasithe case q = 1 · q = 1 holi
Explanation
We write S_n for the sum of the first n terms of a geometric progression: S_n = b₁ + b₁q + b₁q² + ... + b₁q^(n−1). Multiplying both sides by q gives qS_n = b₁q + b₁q² + ... + b₁q^n. Subtracting the first equality from the second cancels almost all the terms, leaving qS_n − S_n = b₁q^n − b₁. For q ≠ 1 this gives S_n = b₁(1 − q^n) : (1 − q) = b₁(q^n − 1) : (q − 1). If q = 1, every term equals b₁, so S_n = b₁ · n. When b_n is known, the sum can also be written S_n = (b_n · q − b₁) : (q − 1).
Worked examples
2 + 6 + 18 + 54 + 162: b₁ = 2, q = 3, n = 5. S₅ = 2 · (3⁵ − 1) : (3 − 1) = 2 · 242 : 2 = 242.
Let the sum 1 + 2 + 4 + ... equal 255. Here b₁ = 1, q = 2: (2^n − 1) : (2 − 1) = 255, so 2^n = 256 = 2⁸ and n = 8.
Class activity
“Group sums”: the class splits into three groups; each computes S_n for its own b₁, q, n; the results are then checked in a table.
Practice
1
If b₁ = 4, q = 2, n = 6, find S₆.
252
2
What is S_n when q = 1?
S_n = b₁ · n
3
Find 1 + 3 + 9 + 27 + 81.
121
4
Why can the formula S_n = b₁(q^n − 1) : (q − 1) not be used when q = 1?
The denominator q − 1 becomes 0, and division by zero is impossible.