Summation Formulas Cheat Sheet

The closed-form formula for the sum of the first n integers is n(n+1)/2. For n=10, that gives 55. Summation formulas for arithmetic series, geometric series, and sums of powers, including the sum of squares formula and sum of cubes formula, are listed below. Each entry includes the closed form, a worked example at n=10, and the conditions for its use. Choose the series type you have, verify the starting index, and apply the correct closed form.

The sum of the first n natural numbers, written Σ_{k=1}^{n} k, has the closed form n(n+1)/2. This is the simplest and most reused summation formula in calculus and discrete mathematics. It appears in OpenStax Calculus Vol. 1 section 5.1 as a standard result.

Short Derivation (Gauss Pairing)

Pair the first and last terms: 1 + n, 2 + (n-1), 3 + (n-2), and so on. Each pair sums to n+1. There are n/2 such pairs when n is even. The total is (n/2)(n+1). The same formula works for odd n because the middle term is (n+1)/2 and the pairing logic still holds. The anecdote that a young Gauss summed 1 to 100 as 50×101 = 5050 is widely repeated, though the earliest source for the story is from 1856, according to Brian Hayes in American Scientist (2006). The method itself is sound for any arithmetic series.

Example at n=10: 10×11/2 = 55. The terms 1+2+3+4+5+6+7+8+9+10 = 55.

Failure case: If the lower bound is not 1, the formula does not apply directly. See the section on adjusting a formula when the index does not start at 1.

Summation formulas for higher powers are polynomial in n. OpenStax Calculus Vol. 1 section 5.1 gives the standard closed forms.

Sum of Squares: Σk² = n(n+1)(2n+1)/6

Verified at n=10: 10×11×21/6 = 2310/6 = 385. The squares 1+4+9+16+25+36+49+64+81+100 = 385.

Derivation method: Use the identity (k+1)³ − k³ = 3k²+3k+1. Sum both sides from k=1 to n; the left side telescopes to (n+1)³ − 1. The right side becomes 3Σk² + 3n(n+1)/2 + n. Solve for Σk².

Sum of Cubes: Σk³ = [n(n+1)/2]²

Verified at n=10: (10×11/2)² = 55² = 3025. The cubes 1+8+27+64+125+216+343+512+729+1000 = 3025.

The elegance of the sum of cubes formula is that the sum equals the square of the sum of integers. This identity is not a coincidence but follows from a telescoping sum involving (k+1)⁴ − k⁴.

For Σk⁴, the closed form is a degree-5 polynomial. A commonly quoted version is n(n+1)(2n+1)(3n²+3n-1)/30. Wolfram MathWorld's 'Power Sum' entry gives the general Faulhaber formula using Bernoulli numbers, which is the authoritative reference for any exponent p.

Verified at n=10: 10×11×21×(300+30-1)/30 = 2310×329/30 = 759990/30 = 25333. The actual sum of fourth powers 1⁴+2⁴+...+10⁴ is 1+16+81+256+625+1296+2401+4096+6561+10000 = 25333. That matches.

The Faulhaber formula generalises this pattern. For any positive integer p, Σ_{k=1}^{n} k^p is a polynomial in n of degree p+1 with rational coefficients expressed using Bernoulli numbers B_j. Wolfram MathWorld gives the form Σ_{k=1}^{n} k^p = 1/(p+1) Σ_{j=0}^{p} (-1)^j C(p+1, j) B_j n^{p+1-j}, where B₀ = 1, B₁ = 1/2. This is the polynomial that produces the sum of squares formula through sum of fourth powers. For p=4, B₂=1/6, B₃=0, B₄=-1/30 produce the cubic factor 3n²+3n-1.

An arithmetic series adds terms that increase by a constant common difference d. The first term is a, and the nth term is a + (n-1)d. Two equivalent formulas give the sum.

Form with d: S = n/2 × [2a + (n-1)d]

Example: Sum the first 10 terms of 3, 7, 11, 15, ... (a=3, d=4). S = 10/2 × [6 + 9×4] = 5 × [6 + 36] = 5 × 42 = 210. Check the first few terms: 3+7+11+15+19+23+27+31+35+39 = 210.

Form with first and last: S = n/2 × (first + last)

Example: Same series, tenth term is 39. S = 10/2 × (3 + 39) = 5 × 42 = 210. This form is simpler but requires knowing the last term.

Failure case: If the sequence is not arithmetic (differences not constant), both formulas give the wrong sum. Check the common difference before applying.

A geometric series multiplies each term by a constant common ratio r. The first term is a. The finite geometric series closed form is S = a(1 − r^n)/(1 − r) for r ≠ 1. OpenStax Calculus Vol. 2 section 5.2 gives this as Σ_{k=0}^{n-1} ar^k = a(1-r^n)/(1-r).

Example: Sum 3, 6, 12, 24, ... for 5 terms (a=3, r=2, n=5). S = 3(1 − 2⁵)/(1 − 2) = 3(1 − 32)/(−1) = 3(−31)/(−1) = 93. Check: 3+6+12+24+48 = 93.

The r = 1 Case

When r = 1, the geometric series closed form produces division by zero. Every term is a, so the sum is simply n × a. For the example above, if r=1 and a=3, five terms sum to 5 × 3 = 15. This is the correct behaviour, not an error.

Failure case: If r is 1 and you use the standard closed form without handling the division by zero, any calculator or code will return an error or undefined result. Always check r ≠ 1 before applying the main closed form.

Most closed-form formulas assume the lower bound is 1. If the sum starts at k = m (m > 1), subtract the sum of terms from 1 to m-1 from the sum of terms from 1 to n. This uses the summation property Σ_{i=1}^{n} a_i = Σ_{i=1}^{m-1} a_i + Σ_{i=m}^{n} a_i (OpenStax Calculus Vol. 1 section 5.1).

Example: Σ_{k=5}^{10} k². Compute Σ_{k=1}^{10} k² = 385 (from earlier). Compute Σ_{k=1}^{4} k² = 4×5×9/6 = 180/6 = 30. Result: 385 − 30 = 355. Check: 25+36+49+64+81+100 = 355.

Failure case: If you forget to adjust the number of terms in an arithmetic or geometric series, you may apply the closed form with n equal to the upper bound instead of the actual term count. The number of terms from m to n inclusive is n − m + 1.

Sum of powers (Σk^p) and power series (Σaₙxⁿ) are different mathematical objects that share the word 'power'. A sum of powers is a finite polynomial in n. Its value is a single number for a given n. A power series is an infinite function of x, written Σaₙxⁿ, that may converge or diverge. The confusion is common but can be avoided by checking the index: if the index appears only in the base (k^p) and the exponent is a constant, it is a sum of powers. If the index appears in the exponent (xⁿ) or multiplies a variable, it is a power series.

Sums of powers are covered here. Infinite geometric series and Taylor series belong elsewhere.

Sum	Closed-Form Formula	Value at n = 10
Σk	n(n+1)/2	55
Σk²	n(n+1)(2n+1)/6	385
Σk³	[n(n+1)/2]²	3025
Σk⁴	n(n+1)(2n+1)(3n²+3n-1)/30	25333
Arithmetic: a=3, d=4	n/2 × [2a + (n-1)d]	210
Geometric: a=3, r=2	a(1-rⁿ)/(1-r)	93

High-school precalculus students who need to evaluate finite sums by hand using closed-form formulas will find every standard summation formula here. College calculus students using sigma notation to define Riemann sums and the definite integral should study the sum of squares formula and sum of cubes formula to evaluate limits of Riemann sums. College discrete-mathematics students working with sums of sequences and combinatorial identities need the arithmetic series formula and geometric series formula for algorithm analysis and counting problems. Teachers checking student work on textbook summation problems can verify results against the table.

Skip this if you need the sum of an infinite series. That subject requires convergence tests, Taylor series, and Fourier series, which are not covered here. Also skip if you need a general-purpose arithmetic calculator; a spreadsheet application handles that directly with less risk of misapplying formulas.