3.9 KiB
title |
---|
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
----------------------------------------
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.
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.