Take any odd number and square it. It will invariably be a multiple of 8 plus 1. So (odd)^2=8n+1 where n is an integer. Show why this is always so. Also show what the pattern for n is.
All odd numbers are of the form 4x+1 or 4x+3. We have two cases.
(4x+1)^2=16x^2+8x+1=8(2x^2+x)+1
n=2x^2+x
(4x+3)^2=16x^2+24x+9=8(2x^2+3x+1)+1
n=2x^2+3x+1
The pattern for n is triangular numbers.
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Posted by Math Man
on 2013-06-10 16:51:54 |
Answer
