Talk:Gnomon (figure)
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[edit] Figure wanted
As a not-very-mathematical (nor very visual) user, I had a time understanding the relationship between the gnomon and the parallelogram. Does anyone have a figure or illustration? Thanks --Ben 11:43, 19 April 2007 (UTC)
[edit] factoring property
Has anyone else noticed that our (recursively) thus far known gnomon (n^2), increased by one "segment" (+2n+1), factors to ((n+1)^2)? Therefore, (n+1)^2=(n+1)(n+1)=1n^2+2n+1. This works for any other exponent too:
(n+1)^3=(n+1)(n+1)(n+1)=1n^3+3n^2+3n+1
(n+1)^4=(n+1)(n+1)(n+1)(n+1)=1n^4+4n^3+6n^2+4n+1
(n+1)^5=(n+1)(n+1)(n+1)(n+1)(n+1)=1n^5+5n^4+10n^3+10n^2+5n+1
(n+1)^6=(n+1)(n+1)(n+1)(n+1)(n+1)(n+1)=1n^6+6n^5+15n^4+20n^3+15n^2+6n+1...
Also, if you examine the coefficients of the expanded forms(Bold), they digitally add to the respective exponent of eleven(Italic)(above):
11^2=121
11^3=1331
11^4=14641
11^5=161051
11^6=1771561 —Preceding comment added by varka (talk) 05:31, 7 March 2008 (UTC)
- See Pascal's triangle --Oni Lukos ct 06:13, 7 March 2008 (UTC)
