Sum of a Linear Number Sequence Calculator
Our free 3d geometry calculator solves sum alinear number sequence problems. Get worked examples, visual aids, and downloadable results.
Formula
S = sum of (a + i*d)^p for i = 0 to n-1
Where a is the first term, d is the common difference, n is the number of terms, and p is the power applied to each term. For p=1 this reduces to the standard arithmetic series sum. For p=2, the sum-of-squares formula applies.
Worked Examples
Example 1: Sum of Squares of Arithmetic Sequence
Problem: Find the sum of squares of the first 5 terms of the sequence 3, 7, 11, 15, 19 (a=3, d=4, n=5, p=2).
Solution: Terms: 3, 7, 11, 15, 19\nSquares: 9, 49, 121, 225, 361\nSum of squares = 9 + 49 + 121 + 225 + 361 = 765\nVerification with formula: n*a^2 + n(n-1)*a*d + d^2*n(n-1)(2n-1)/6\n= 5(9) + 5(4)(3)(4) + 16(5)(4)(9)/6\n= 45 + 240 + 480 = 765
Result: Sum of squares = 765 | Linear sum = 55 | Mean squared term = 153
Example 2: Sum of Cubes of Natural Numbers
Problem: Calculate the sum of cubes of the first 6 natural numbers (a=1, d=1, n=6, p=3).
Solution: Terms: 1, 2, 3, 4, 5, 6\nCubes: 1, 8, 27, 64, 125, 216\nSum of cubes = 1 + 8 + 27 + 64 + 125 + 216 = 441\nVerification: [n(n+1)/2]^2 = [6(7)/2]^2 = 21^2 = 441
Result: Sum of cubes = 441 = 21^2 | Linear sum = 21 | Confirms Nicomachus theorem
Frequently Asked Questions
What is an alinear number sequence and how does it differ from a linear one?
An alinear (or nonlinear) number sequence is formed by applying a nonlinear transformation, such as raising to a power, to the terms of an arithmetic (linear) sequence. In a standard linear arithmetic sequence like 2, 5, 8, 11, 14, each term increases by a constant difference. An alinear transformation might square each term to produce 4, 25, 64, 121, 196, where the differences between consecutive terms are no longer constant. The sum of such transformed sequences cannot be computed using the simple arithmetic series formula n/2 times (first + last) and instead requires either direct computation or specialized formulas depending on the power applied.
How do you calculate the sum of squares of an arithmetic sequence?
For an arithmetic sequence with first term a, common difference d, and n terms, the sum of squares has a closed-form formula. It equals n times a squared, plus n times (n-1) times a times d, plus d squared times n times (n-1) times (2n-1) divided by 6. This formula derives from expanding (a + id)^2 = a^2 + 2aid + i^2d^2 and summing each part separately using known formulas for sum of integers and sum of squares of integers. For the basic sequence 1, 2, 3, ..., n (where a=1 and d=1), the sum of squares simplifies to the well-known formula n(n+1)(2n+1)/6. This result has applications in statistics for computing variance.
What is the formula for the sum of an arithmetic series?
The sum of a standard arithmetic series with first term a, common difference d, and n terms is S = n/2 times (2a + (n-1)d), which can also be written as S = n/2 times (first term + last term). This formula was famously discovered by young Carl Friedrich Gauss, who reportedly summed the numbers 1 to 100 by pairing 1 with 100, 2 with 99, and so on, yielding 50 pairs each summing to 101, for a total of 5050. The formula works because the average of an arithmetic sequence is exactly the average of its first and last terms. This linear sum serves as the baseline from which alinear sums deviate.
What are Faulhaber formulas and how do they relate to alinear sums?
Faulhaber formulas provide closed-form expressions for the sum of the p-th powers of the first n natural numbers: 1^p + 2^p + 3^p + ... + n^p. For p=1, the sum is n(n+1)/2. For p=2, it is n(n+1)(2n+1)/6. For p=3, it is [n(n+1)/2]^2, which is remarkably the square of the sum of the first n numbers. For higher powers, the formulas involve Bernoulli numbers and become increasingly complex. These formulas can be adapted to arithmetic sequences with arbitrary first term and common difference by substituting the general term a + id for each i. Faulhaber published these results in 1631, and they remain important in number theory and combinatorics.
How is the sum of cubes related to the sum of natural numbers?
One of the most elegant identities in mathematics states that the sum of cubes of the first n natural numbers equals the square of the sum of those numbers. Specifically, 1^3 + 2^3 + 3^3 + ... + n^3 = [n(n+1)/2]^2. For example, 1^3 + 2^3 + 3^3 + 4^3 = 1 + 8 + 27 + 64 = 100, and (4 times 5 / 2)^2 = 10^2 = 100. This identity, sometimes called Nicomachus theorem after the ancient Greek mathematician, can be proven by mathematical induction or by geometric argument. For general arithmetic sequences, the sum of cubes formula is more complex and involves terms up to the fourth power of n.
Can you compute the sum of fractional powers of a sequence?
Yes, you can compute the sum of fractional powers such as square roots or cube roots of arithmetic sequence terms, but there are no simple closed-form formulas for most fractional powers. Unlike integer powers where Faulhaber formulas exist, fractional powers require either direct term-by-term computation or numerical approximation techniques. For example, summing sqrt(1) + sqrt(2) + sqrt(3) + ... + sqrt(n) can be approximated by the integral of sqrt(x) from 0 to n, which equals (2/3)n^(3/2), but the exact sum differs from this approximation by an amount related to the Euler-Maclaurin formula. Sum of a Linear Number Sequence Calculator handles fractional powers through direct computation of each term.