Mathematica Mathematical Induction

  • Thread starter 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 doesnt 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 hw 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.
 

matt grime

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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.
 

HallsofIvy

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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 doesnt 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 hw 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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