Evens and Square Numbers
Prove that adding up even numbers starting with 0 cannot addup to a square number
Ex. 0+2+4+6+8+...
OR
Show the square number it adds up to and its root
- Jordan McMichael
Ex. 0+2+4+6+8+...
OR
Show the square number it adds up to and its root
- Jordan McMichael
Labels: SharedPuzzle
posted by ToM at
8:28 PM
6 Comments:
0+2+4+6+8+...
= 2(0+1+2+3+4+5+...)
The partial sums inside parentheses are
0,1,3,6,10,15,21,28 ...
which are the triangular numbers.
The formula for the n'th triangular number is
n(n-1)/2
Twice this is just
n(n-1)
This number can never be a perfect square.
The nth square number is 1 + 3 + 5 + 7 + 9 + ... + (2n-1) = n²,
where n is a natural number. The series contains n terms.
Now add 1 to each term
=> (1+1) + (3+1) + (5+1) + ... +((2n-1)+1) = n² + n
=> 2 + 4 + 6 + ... + 2n = n(n+1).
Require n(n+1) = m² => m is the geometric mean of n and n+1
and lies in between them. As there is no natural number between
n and n+1 this isn't possible.
For completeness:
n(n+1) = n²(1 + 1/n) > n²
n(n+1) = (n+1)²(1 - 1/(n+1)) < (n+1)²
Let m² = n(n+1), then (n+1) > m > n
This post has been removed by the author.
I do believe you got it Chris
-Jordan
Thanks Jordan. But Ross had beaten me to it.
Also my last comment is clearer as:
(n+1)² > n(n+1) > n²
Hence (n+1) > √(n(n+1)) > n, for any natural number, n.
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