From 14a89d5a56ae0b32cb8ef7c682aed275954bbff5 Mon Sep 17 00:00:00 2001 From: Alexander Molnar <37451552+BTmathic@users.noreply.github.com> Date: Fri, 2 Nov 2018 12:39:56 -0400 Subject: [PATCH] Added content to stub (#34168) Explained what an arithmetic series is, gave the formula for the sum (when finite) and gave two proofs (not just induction) as the 'famous' method by Gauss is too elegant to ignore. --- .../index.md | 93 ++++++++++++++++++- 1 file changed, 88 insertions(+), 5 deletions(-) diff --git a/guide/english/mathematics/proof-of-finite-arithmetic-series-formula-by-induction/index.md b/guide/english/mathematics/proof-of-finite-arithmetic-series-formula-by-induction/index.md index 08e796473a..0a0ca2645e 100644 --- a/guide/english/mathematics/proof-of-finite-arithmetic-series-formula-by-induction/index.md +++ b/guide/english/mathematics/proof-of-finite-arithmetic-series-formula-by-induction/index.md @@ -3,13 +3,96 @@ title: Proof of Finite Arithmetic Series Formula by Induction --- ## Proof of Finite Arithmetic Series Formula by Induction -This is a stub. Help our community expand it. +An arithmetic *sequence* is a sequence of numbers with every consecutive pair having the same difference. For example, +``` +1, 4, 7, 10 +``` +is an arithmetic sequence because `4 - 1 = 3`, `7 - 4 = 3` and `10 - 7 = 3`. An arithmetic *series* is the sum of an arithmetic sequence, for example +``` +1 + 4 + 7 + 10. +``` +The sum of an infinite arithmetic series is not a number, so the question 'what is the value of an arithmetic series' is only interesting for finite arithmetc sequences/series, so we only focus on these here. -This quick style guide will help ensure your pull request gets accepted. +The positive integers up to (and including) 100 is another arithmetic sequence, and we can ask what the corresponding series is, i.e., what is +``` +1 + 2 + 3 + ... + 98 + 99 + 100? +``` +Famously, [Carl Friedrich Gauss](https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss) solved this problem at the age of 7 faster than anyone else in his class by noting a pattern, +``` + 1 + 2 + 3 + ... + 98 + 99 + 100 +100 + 99 + 98 + ... + 3 + 2 + 1 +---------------------------------------- +101 + 101 + 101 + ... + 101 + 101 + 101 +``` +To find the sum we can take twice the sum instead and notice that every opposite pair `(1, 100), (2, 99), etc...` has the same sum `101`, with `100` terms in the series, so adding two coipes of the series gives `100*101`, then dividing by two to get the sum of the original series, the value is +``` +100*101/2. +``` +This idea immediately generalizes to showing +``` +1 + 2 + 3 + ... + n = n*(n + 1)/2 +``` +for any positive integer `n`, as there are `n` terms pairing up with sums `n + 1`. - +But our first example above did not start at 1, nor increment by 1, so what can we do here? If we simply shift the sequence to start at `a` instead of 1 then we have +``` + a + (a + 1) + ... + (a + (n-1)) + (a + n) + (a + n) + (a + (n-1) + ... + (a + 1) + a +------------------------------------------------------ +(2*a + n) + (2*a + n) + ... + (2*a + n) + (2*a + n) +``` +and its sum can be read off easily once again, as +``` +(2*a + n)*n/2. +``` +Lastly, if instead of incrementing by 1 with each term we increment by some value `k`, the same trick can be used and we see that an arithmetic series +``` +a + (a + k) + (a + 2*k) + (a + 3*k) + ... + (a + n*k) +``` +that starts at `a`, increases by `k` each term and has `n+1` terms has the value +``` +(2*a + n*k)*(n + 1)/2. +``` -#### More Information: - +However, in case one feels like this is just a trick and prefers algebraic manipulation, we can also prove this formula by [mathematical induction](https://en.wikipedia.org/wiki/Mathematical_induction). +The base case with `n = 0` is clear, the sum of the series +``` +a +``` +is +``` +a = (2*a + 0)*(0 + 1)/2, +``` +so our formula is correct for the base case. +For the inductive hypothesis, suppose we have a series +``` +a + (a + k) + (a + 2*k) + (a + 3*k) + ... + (a + (n-1)*k) +``` +with value given by the formula, +``` +(2*a + (n-1)*k)*n/2. +``` +Then +``` + a + (a + k) + (a + 2*k) + (a + 3*k) + ... + (a + (n-1)*k) + (a + n*k) += [a + (a + k) + (a + 2*k) + (a + 3*k) + ... + (a + (n-1)*k)] + (a + n*k) += (2*a + (n-1)*k)*n/2 + (a + n*k) +``` +by the induction hypothesis, and this can be simplified as follows +``` +(2*a + (n-1)*k)*n/2 + (a + n*k) = [(2*a + (n-1)*k)*n + 2*a + 2*n*k]/2 (common denominator) + = [2*a*n + (n-1)*n*k + 2*a + 2*n*k]/2 (expanding brackets) + = [(2*a*n + 2*a) + (n^2 - n + 2*n)*k]/2 (collecting like terms) + = [2*a*(n + 1) + n*(n + 1)*k]/2 (simplifying) + = (2*a + n*k)*(n + 1)/2. (factoring the (n + 1)) +``` +Thus, by the principal of mathematical induction, the sum of the arithmetic series +``` +a + (a + k) + (a + 2*k) + (a + 3*k) + ... + (a + (n-1)*k) + (a + n*k) +``` +is indeed the formula given above, +``` +(2*a + n*k)*(n + 1)/2. +```