Arithmetico-Geometric Series
On this page
Direct answer
1 + 3/4 + 5/16 + 7/64 + ... is neither arithmetic nor geometric, yet one subtraction collapses it: multiply by the common ratio 1/4, shift the alignment, and subtract. That manoeuvre defines every arithmetico-geometric progression (AGP), a series whose terms have the form (a + (r−1)d)·x^(r−1) — an arithmetic part riding a geometric part. For the infinite case with |x| < 1, S∞ = a/(1 − x) + dx/(1 − x)². For finite sums the same subtraction delivers a closed form, and it works even when |x| ≥ 1, where no infinite sum exists.
What you must remember
- Method over memory: form S − xS after multiplying by the common ratio; the difference is a clean geometric series in d alone — solve the linear equation for S.
- Infinite sum: S∞ = a/(1 − x) + dx/(1 − x)² for |x| < 1; the series 1 + 3/4 + 5/16 + ... has a = 1, d = 2, x = 1/4, giving 4/3 + 8/9 = 20/9.
- Pure weighted form: Σ_(r=1)^n r·x^(r−1) = [1 − (n+1)x^n + n·x^(n+1)]/(1 − x)² — the derivative of the finite geometric sum.
- Standard infinite values: Σ(r=1)^∞ r/2^r = 2 and Σ(r=1)^∞ r·x^r = x/(1 − x)² for |x| < 1 — quotable on sight.
- Convergence guard: the infinite formula needs |x| < 1 strictly; with |x| ≥ 1 only partial sums exist, and questions exploit exactly this distinction.
- Alternating cases: x = −1/2 or x = −1/3 turn the series alternating; the same subtraction works with sign care in the alignment step.
- Expected-value link: for a geometric-type distribution P(X = r) = q^(r−1)p, the expectation E(X) is an AGP sum — the same subtraction computes it.
One subtraction, one answer
Compute S = 1 + 3/4 + 5/16 + 7/64 + ... completely. Multiply by 1/4: (1/4)S = 1/4 + 3/16 + 5/64 + ... Now subtract term-aligned — the constant 1 stands alone, and each surviving coefficient drops by 2: S − (1/4)S = 1 + 2(1/4 + 1/16 + 1/64 + ...). The bracket is a geometric series with first term 1/4 and ratio 1/4, summing to (1/4)/(3/4) = 1/3. So (3/4)S = 1 + 2/3 = 5/3, and S = 20/9. The two disciplines worth keeping: align powers before subtracting (a misalignment invents an extra term), and name the a, d, x values so the formula version gives the same 20/9 as a cross-check.
Now the finite cousin with a forbidden ratio: sum 1·2 + 2·2² + 3·2³ + ... + n·2^n. Here x = 2, so no infinite sum exists, but the subtraction survives: 2S shifts every term, and S − 2S = 2 + 2² + ... + 2^n − n·2^(n+1) = 2(2^n − 1) − n·2^(n+1). Solving, S = (n − 1)·2^(n+1) + 2. Verify at n = 1: 2 = 0 + 2 ✓; at n = 2: 2 + 8 = 10 = 8 + 2 ✓. The moral: the method is valid for every x; only the infinite limit demands |x| < 1.
Exam patterns and traps
Main loves the infinite AGP with x = 1/2, 1/3, 1/4 or −1/2 in numerical-value questions, plus the one-liner Σ r/2^r = 2; Advanced dresses the same sum inside probability (expected values with geometric-type probabilities) or asks finite sums with |x| > 1 where the infinite formula sits in the options as bait. The standing traps: deploying S∞ when the question asks the nth partial sum; misaligning terms during subtraction, which corrupts the geometric tail; and differentiating Σ x^r for infinite sums without noting the |x| < 1 licence that legitimises termwise differentiation. Reading whether the question says "sum to infinity" or "sum of n terms" decides the correct branch before any algebra starts.
Frequently asked questions
What is the sum of 1 + 2/3 + 3/9 + 4/27 + ...?
An AGP with a = 1, d = 1, x = 1/3: S∞ = 1/(2/3) + (1/3)/(4/9) = 3/2 + 3/4 = 9/4.
When is the infinite AGP formula valid?
Only for |x| < 1; beyond that the geometric tail diverges and the series has no finite sum.
How do I sum 1·2 + 2·2² + 3·2³ + ... + n·2^n?
Multiply by 2 and subtract: the answer is (n − 1)·2^(n+1) + 2 — the subtraction works for finite sums even when |x| > 1.
What is Σ_(r=1)^∞ r/2^r?
2 — the most quoted infinite AGP value, from Σ r·x^r = x/(1 − x)² at x = 1/2.
How is AGP linked to expectation?
For a geometric-type distribution P(X = r) ∝ q^(r−1)p, the expectation is an AGP sum, computed by the same multiply-and-subtract routine.