Inductive Proof

Sep 2009
181
0
how do you get this problem \(\displaystyle [k(k+1)/2]^2 + (k+1)^3\) down to \(\displaystyle [(k+1)(k+2)/2]^2\)

I'm having troubles with the arithmetic...
 
Nov 2009
485
184
Wait, is this true?

Factor \(\displaystyle (k+2)^2\) out to get \(\displaystyle (k+1)^2\left(\frac{k}{2}+(k+1)\right)\). Then combine the terms by getting a common denominator. But, it doesn't seem to give you what claim it does.
 
Sep 2009
181
0
so i factored out (k+1)^2 and i got this....1/4(k+1)^2 + (K^2+(k+1))
 
Nov 2009
485
184
Ah, oops. I factored it incorrectly, myself.

You should have \(\displaystyle
(k+1)^2\left(\frac{k^2}{4}+(k+1)\right)=(k+1)^2\left(\frac{k^2+4k+4}{4}\right)=(k+1)^2\frac{(k+2)^2}{4}
\)

This gives it to you.
 
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