99 lines
3.9 KiB
Markdown
99 lines
3.9 KiB
Markdown
---
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title: Proof of Finite Arithmetic Series Formula by Induction
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---
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## Proof of Finite Arithmetic Series Formula by Induction
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An arithmetic *sequence* is a sequence of numbers with every consecutive pair having the same difference. For example,
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```
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1, 4, 7, 10
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```
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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
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```
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1 + 4 + 7 + 10.
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```
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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.
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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
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```
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1 + 2 + 3 + ... + 98 + 99 + 100?
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```
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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,
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```
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1 + 2 + 3 + ... + 98 + 99 + 100
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100 + 99 + 98 + ... + 3 + 2 + 1
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----------------------------------------
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101 + 101 + 101 + ... + 101 + 101 + 101
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```
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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
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```
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100*101/2.
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```
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This idea immediately generalizes to showing
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```
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1 + 2 + 3 + ... + n = n*(n + 1)/2
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```
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for any positive integer `n`, as there are `n` terms pairing up with sums `n + 1`.
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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
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```
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a + (a + 1) + ... + (a + (n-1)) + (a + n)
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(a + n) + (a + (n-1) + ... + (a + 1) + a
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------------------------------------------------------
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(2*a + n) + (2*a + n) + ... + (2*a + n) + (2*a + n)
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```
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and its sum can be read off easily once again, as
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```
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(2*a + n)*n/2.
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```
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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
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```
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a + (a + k) + (a + 2*k) + (a + 3*k) + ... + (a + n*k)
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```
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that starts at `a`, increases by `k` each term and has `n+1` terms has the value
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```
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(2*a + n*k)*(n + 1)/2.
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```
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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).
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The base case with `n = 0` is clear, the sum of the series
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```
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a
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```
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is
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```
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a = (2*a + 0)*(0 + 1)/2,
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```
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so our formula is correct for the base case.
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For the inductive hypothesis, suppose we have a series
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```
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a + (a + k) + (a + 2*k) + (a + 3*k) + ... + (a + (n-1)*k)
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```
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with value given by the formula,
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```
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(2*a + (n-1)*k)*n/2.
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```
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Then
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```
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a + (a + k) + (a + 2*k) + (a + 3*k) + ... + (a + (n-1)*k) + (a + n*k)
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= [a + (a + k) + (a + 2*k) + (a + 3*k) + ... + (a + (n-1)*k)] + (a + n*k)
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= (2*a + (n-1)*k)*n/2 + (a + n*k)
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```
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by the induction hypothesis, and this can be simplified as follows
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```
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(2*a + (n-1)*k)*n/2 + (a + n*k) = [(2*a + (n-1)*k)*n + 2*a + 2*n*k]/2 (common denominator)
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= [2*a*n + (n-1)*n*k + 2*a + 2*n*k]/2 (expanding brackets)
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= [(2*a*n + 2*a) + (n^2 - n + 2*n)*k]/2 (collecting like terms)
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= [2*a*(n + 1) + n*(n + 1)*k]/2 (simplifying)
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= (2*a + n*k)*(n + 1)/2. (factoring the (n + 1))
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```
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Thus, by the principal of mathematical induction, the sum of the arithmetic series
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```
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a + (a + k) + (a + 2*k) + (a + 3*k) + ... + (a + (n-1)*k) + (a + n*k)
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```
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is indeed the formula given above,
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```
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(2*a + n*k)*(n + 1)/2.
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```
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