Parity
http://mathworld.wolfram.com/Parity.html
Let be a positive integer and
be any integer with the same parity as
with
These odd integers are all positive.
for example,
If n = 2, m = 4
8 = 3 + 5
the product is equal to the sum of 2 (n = 2) consecutive odd integers
If n = 3, m = 5
15 = 3 + 5 + 7
the product is equal to the sum of 3 (n = 3) consecutive odd integers
If n = 5, m = 7
35 = 3 + 5 + 7 + 9 + 11
the product is equal to the sum of 5 (n = 5) consecutive odd integers
Can you prove the product will always equal to the sum of
consecutive odd integers?
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Proof –
The sum of the first odd integers is
:
From this, we have that the sum of any sequence of consecutive odd integers is :
So an integer can be written as the sum of consecutive odd integers if it can be expressed as
where
and
or
or
is odd,
is odd :
If
The first odd integer in the sequence is: 2(1) + 1 = 3
The last odd integer in the sequence is 2(4) – 1 = 7
We have the solution
is even,
is even :
E.g.
a = 12, b = 6
n = (a+b)/2 n = (12+6)/2 = 9
m = (a-b)/2 m = (12-6)/2 = 3
1st odd integer is : 2(3+1) – 1 = 7
last odd integer is : 2(9) – 1 = 17
We have the solution 7 + 9 + 11 + 13 + 15 + 17 = 72