Problem: Find the sum of the first 1000 natural numbers.
Solution: Write the numbers twice as follows:
Summing the numbers in each column, we have:
(1000 times)
And dividing by two we have
This method clearly works not just for 1000, but any natural number .
Problem: Find the sum of the squares of the first natural numbers
Solution: Arrange the numbers three times, in triangles as follows:
Here we prove what is obvious from the picture: that the numbers in the “sum” triangle work out as expected.
In the leftmost triangle, there are copies of
in the
th row (indexing from 1), so the sum of the
th row is
, and the sum of all the numbers in the triangle is
. Since the other two “summand” triangles are just rotations of the first triangle, their sums are also the sum of the first
squares.
In the “sum” triangle at the th row
spaces from the left (indexing both from zero), we have the value
. In each triangle there are
different numbers (as per the first proof), so there are that many copies of
in the fourth triangle. Summing up the fourth triangle gives three times our desired result, and hence
, or
This second proof was shown to me by Sándor Dobos, a professor of mine in Budapest whom I remember fondly.
I like that , and thx that is so cool
LikeLike
Pingback: Methods of Proof — Induction | Math ∩ Programming
Pingback: Elliptic Curves as Elementary Equations | Math ∩ Programming
Pingback: Deconstructing the Common Core Mathematical Standard | Math ∩ Programming