Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.


paul c
Posts:
3
Registered:
12/6/04


[No Subject]
Posted:
Mar 21, 2000 4:38 AM


denise,
you need to show the following steps:
n(n+1)(n+2)(n+3)/4 + (n+1)(n+2)(n+3) = [n(n+1)(n+2)(n+3) + 4(n+1)(n+2)(n+3)]/4
factoring (n+1)(n+2)(n+3) in the numerator, we get the desired expression (n+1)(n+2)(n+3)(n+4)/4.
paul cabral assistant software engineer adtexphilippines, inc.
p.s. enjoy diversity in all forms!
From: Denise <DCamillo@aol.com>  Block address To: discretemath@forum.swarthmore.edu Date: Sun, 19 Mar 2000 16:57:41 +0000 Subject: Mathematical Induction Add Addresses
Need to show by induction: 1*2*3+2*3*4+...n*(n+1)*(n+2)=n*(n+1)*(n+2)*(n+3)/4. Please help!!!!!! Email me at DCamillo@aol.com, Thanks.



