Summation Calculator
Evaluate a sigma summation of any expression in i, from a start index to an end index, with each expanded term shown.
Summation Calculator
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Formula: Σ from i=start to end of f(i) = f(start) + f(start+1) + ... + f(end)
Worked example — 338,350
Formula
Σ from i=start to end of f(i) = f(start) + f(start+1) + ... + f(end)
Sigma notation compactly represents the sum of a series. The expression is evaluated for each integer value of i from start to end. Common formulas: Σi = n(n+1)/2, Σi² = n(n+1)(2n+1)/6.
Worked Examples
Example 1: Sum of squares 1 to 100
Problem:Evaluate the sum of i squared for i from 1 to 100.
Solution:Use the closed form for the sum of the first n squares: n(n + 1)(2n + 1)/6. With n = 100: 100 x 101 x 201 / 6 = 2,030,100 / 6 = 338,350 Sanity check against the average term: the mean square is 338,350/100 = 3,383.5, which sits between 1 and 10,000 as expected for squares up to 100 squared.
Result:338,350
Example 2: Arithmetic sum with the Gauss pairing trick
Problem:Evaluate the sum of i for i from 1 to 200, then the sum of (3i + 2) over the same range.
Solution:Sum of i = n(n + 1)/2 = 200 x 201 / 2 = 20,100 Pairing check: 1 + 200 = 201, and there are 100 such pairs, so 100 x 201 = 20,100. For the sum of (3i + 2), split the sigma by linearity: sum(3i + 2) = 3 x sum(i) + 2 x 200 = 3(20,100) + 400 = 60,300 + 400 = 60,700
Result:sum of i = 20,100 | sum of (3i + 2) = 60,700
Example 3: Finite geometric sum
Problem:Evaluate the sum of 2 to the power i for i from 0 to 10.
Solution:A geometric sum with first term a = 1 and ratio r = 2 has closed form a(r^(n+1) - 1)/(r - 1) for i running from 0 to n. With a = 1, r = 2, n = 10: (2^11 - 1)/(2 - 1) = (2048 - 1)/1 = 2047 Direct check: 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 = 2047. Note the pattern: the sum of powers of 2 up to 2^n is always 2^(n+1) - 1, which is why an 11-bit unsigned register tops out at 2047.
Result:2,047
Frequently Asked Questions
What are common summation formulas?
Σi (1 to n) = n(n+1)/2. Σi² = n(n+1)(2n+1)/6. Σi³ = [n(n+1)/2]². Σaʳ = a(rⁿ-1)/(r-1) for geometric series.
What expressions can I use?
Use JavaScript: i*i for i², Math.pow(i,3) for i³, 1/i for harmonic series, Math.pow(2,i) for geometric, i*(i+1)/2 for triangular numbers.
What is the difference between a finite sum and an infinite series?
A finite summation (Σ from start to end) always produces a definite number when the expression is defined at each integer index. An infinite series (Σ from n=1 to ∞) only converges to a finite value if the terms decrease fast enough. For example, Σ 1/2ⁿ = 1 (converges), but Σ 1/n diverges. Use the Series Convergence calculator to test whether an infinite series converges.
How is sigma notation used in probability and statistics?
In statistics, sigma notation appears in mean (x̄ = Σxᵢ/n), variance (σ² = Σ(xᵢ - x̄)²/n), and standard deviation calculations. In probability, expected value is E[X] = Σ x·P(x) summed over all outcomes. Summation notation compactly expresses operations over all data points or outcomes, making it foundational for statistical formulas involving large datasets.
How do I shift the lower limit of a summation?
Split it: the sum from i = a to b equals the sum from 1 to b minus the sum from 1 to a − 1. For Σ i from 5 to 10 that is 55 − 10 = 45, which matches 5 + 6 + 7 + 8 + 9 + 10 directly. The same subtraction lets you apply the standard closed forms, all of which are written starting at i = 1, to any starting index you like.
What algebraic rules can I apply to sigma notation?
Summation is linear. A constant factor pulls straight out, so Σ c·f(i) = c·Σ f(i); sums split, so Σ (f(i) + g(i)) = Σ f(i) + Σ g(i); and a constant term adds up to n·c over n indices. What you cannot do is move a factor that depends on i outside the sigma, or split a product, because Σ f(i)g(i) is not Σ f(i) times Σ g(i).
What is a telescoping sum?
One in which each term is a difference that cancels against its neighbours, leaving only the two ends. The sum of (1/i − 1/(i+1)) from i = 1 to n collapses to 1 − 1/(n+1), which equals n/(n+1). Partial fractions often expose this structure, and telescoping is one of the few ways to obtain an exact closed form for a sum with no standard formula.
How does a double summation work?
Nested sigmas evaluate from the inside out: the inner index runs a full cycle for every single value of the outer index, so two indices each running 1 to n produce n² terms. When the inner limits do not depend on the outer index, the order of the two sums can be swapped freely, and a separable term f(i)g(j) factors into the product of two single sums.
Why does the harmonic sum grow so slowly?
Σ 1/i has no simple closed form, but it tracks the natural logarithm of n plus a constant near 0.5772, the Euler-Mascheroni constant. Because the logarithm keeps growing, the sum diverges, yet doubling the number of terms adds only about ln 2, roughly 0.69, no matter how far out you already are. That is why it can look convergent in a numerical experiment.
How can I check that a closed-form summation formula is correct?
Test it at small n and then prove it by induction. For Σ i² with n = 3 the direct sum is 1 + 4 + 9 = 14, and n(n+1)(2n+1)/6 gives 3 × 4 × 7 / 6 = 14, so they agree. Induction then extends it to every n: verify the base case, then show that adding the next term reproduces the formula with n + 1 substituted.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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