Mathematica Factor Math Induction: What's the Common Factor?

AI Thread Summary
The discussion centers around factoring the expression [k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)]/4. Participants explore the common factors present in the terms, specifically noting that (k+1), (k+2), and (k+3) are shared. The expression can be factored to yield (k+1)(k+2)(k+3)(k + 4)/4. There is also a suggestion to substitute variables for each term to simplify understanding, but the consensus emphasizes the use of the distributive property rather than mathematical induction for this problem. Overall, the focus is on recognizing and extracting common factors to simplify the expression effectively.
L²Cc
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[k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)]/4
factor this out...
What's the common factor? How did you get there? (ok i hope it doesn't require expanding the polynomials :p)
Again, would it be easier if i substituted every (k+x) by a different variable, where (k+1) would equal to variable 'A', (k+2) = B, and so forth?
(this would eventually lead to a mathematical induction)...this is not a homework question...im simply trying to understand factoring and mathematical induction...
 
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What are you asking? I can't make sense of your post or how it relates to mathematical induction.
 
As far as I can tell all you need to remember is that xy+xz=x(y+z), i.e. the distributive property of multiplication. And it has nothing to do with induction.
 
L²Cc said:
[k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)]/4
factor this out...
What's the common factor? How did you get there? (ok i hope it doesn't require expanding the polynomials :p)
Again, would it be easier if i substituted every (k+x) by a different variable, where (k+1) would equal to variable 'A', (k+2) = B, and so forth?
(this would eventually lead to a mathematical induction)...this is not a homework question...im simply trying to understand factoring and mathematical induction...
It's obvious, isn't it, that there are factors of k+1, k+2, and k+3 in both terms? You can factor those out:
[k(k+1)(k+2)(k+3)+ 4(k+1)(k+2)k+3)]/4= (k+1)(k+2)(k+3)[k+ 4]/4.
 

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