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roam
Sep3-09, 02:36 AM
1. The problem statement, all variables and given/known data

I need some help with this Algebra problem. In the following I don't know how they manipulated the RHS to get to the LHS:

http://img402.imageshack.us/img402/7176/solns.gif

2. Relevant equations

3. The attempt at a solution

\frac{1}{4}(k+1)^2 .k^2 + (k+1)^3

\frac{1}{4}(k+1)^2 .k^2 + k^3 + 3k^2 +3k +1

I don't know how to manipulate k^2 + k^3 + 3k^2 +3k +1 into (k^2+4k+4)
symbolipoint
Sep3-09, 03:05 AM
Are you required to carry steps to change the right-side to be equal to the left-side? You should be allowed to carry steps on both sides so that you can show the right and left sides are equal to a third expression. The idea is, that if a=b, and if b=c, then a=c.
NJunJie
Sep3-09, 03:54 AM
another opinion of mine personally is to resolve 'complex' question initially is to model them by substitution.
Like letting another alegrabic representation (eg) Let a = K+1 and then it will look simpler.

Hope it helps.
roam
Sep3-09, 05:10 AM
Are you required to carry steps to change the right-side to be equal to the left-side? You should be allowed to carry steps on both sides so that you can show the right and left sides are equal to a third expression. The idea is, that if a=b, and if b=c, then a=c.

Well I'm not sure what they've done there. I mean how they simplified [\frac{1}{2}k(k+1)]^2 + (k+1)^3 into \frac{1}{4} (k+1)^2 (k^2+4k+4)

Anyway, here's the rest (it's from a proof by induction problem):

http://img200.imageshack.us/img200/6978/62815040.gif
njama
Sep3-09, 05:29 AM
[\frac{1}{2}k(k+1)]^2 + (k+1)^3=(\frac{1}{2})^2k^2(k+1)^2+(k+1)(k+1)^2

Now just factor (k+1)2 and you are done. :approve:
roam
Sep3-09, 03:48 PM
[\frac{1}{2}k(k+1)]^2 + (k+1)^3=(\frac{1}{2})^2k^2(k+1)^2+(k+1)(k+1)^2

Now just factor (k+1)2 and you are done. :approve:

But that doesn't work! If I factor out the (k+1)2 I will have:

\frac{1}{4}(k+1)^2 (k^2+k+1)

Which is not the same as:

\frac{1}{4} (k+1)^2 (k^2+4k+4)

:uhh:
njama
Sep3-09, 05:31 PM
But that doesn't work! If I factor out the (k+1)2 I will have:

\frac{1}{4}(k+1)^2 (k^2+k+1)

Which is not the same as:

\frac{1}{4} (k+1)^2 (k^2+4k+4)

:uhh:

You are wrong. :yuck:

Do the factorization again.

=(k+1)^2(\frac{1}{4}k^2+k+1)

Now factor 1/4 and see what will you come up with.
roam
Sep3-09, 06:09 PM
Awww! Gee! I see what you mean now! Thanks a lot for the help.