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| Proof of Finite Arithmetic Series Formula by Induction |
Proof of Finite Arithmetic Series Formula by Induction
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.
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 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
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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
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(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.
However, in case one feels like this is just a trick and prefers algebraic manipulation, we can also prove this formula by 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.