Factoring (n+1)! - 1 + (n+1)(n+1)! Step-by-Step

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SUMMARY

The discussion focuses on the step-by-step factoring of the expression (n+1)! - 1 + (n+1)(n+1)!. Participants clarify that the expression can be simplified to (n+1)!*(1 + n + 1) - 1. The conversation emphasizes the importance of understanding factorial notation, specifically that (n+1)! equals (n+1)*n!. Additionally, the discussion touches on the concept of mathematical induction and the potential benefits of substituting variables for clarity in complex expressions.

PREREQUISITES
  • Understanding of factorial notation, specifically (n+1)!
  • Basic algebraic manipulation skills, including factoring expressions.
  • Familiarity with mathematical induction principles.
  • Knowledge of polynomial expressions and their factoring.
NEXT STEPS
  • Study the properties of factorials and their applications in combinatorics.
  • Learn about mathematical induction and its role in proving statements about integers.
  • Explore polynomial factoring techniques, including common factors and substitution methods.
  • Practice simplifying complex algebraic expressions involving factorials and polynomials.
USEFUL FOR

Students and educators in mathematics, particularly those studying algebra, factorials, and mathematical induction, will benefit from this discussion.

L²Cc
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can you please explain (step by step) how to factor the following:

(n+1)! - 1 + (n+1)(n+1)!

i have the answer, don't know how to get there!
 
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radou, that's absolutely wrong. You realized you just proved (n+1)!=1 for all n?
 
Office_Shredder said:
radou, that's absolutely wrong. You realized you just proved (n+1)!=1 for all n?

Yes, I just did. :biggrin:
 
That's not fair! You can't delete your post like that! :P

Getting back on topic:

(n+1)! + (n+1)*(n+1)! - 1 = (n+1)!*(1 + n + 1) - 1.

Can you go from there?
 
thanks for the quick reply...
You see that's where i get confused...how did you end up with (1 + n + 1)...
Is (n+1)! = (n-1)(n)(n+1)...and so forth?!
 
(n+1)! + (n+1)*(n+1)! = (n+1)!*1 + (n+1)!*(n+1). You factor (n+1)! out and are left with 1 + n + 1

And yes, (n+1)! = (n+1)*n*(n-1)...
 
does the factoring process of (n+1)! involve (n+1)! = (n+1)*n*(n-1)... ?
 
(n+1)! = n!(n+1)
 
oh all right i see what you guys are coming at...would it have been easier if i had substituted any variable (say, h) for (n+1)!...? and then factored it...
Btw, this is part of a mathematical induction...im trying to understand factorials better!
thank you guys!
 
  • #10
(this does not involve factorials anymore)...
[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?
 
Last edited:
  • #11
L²Cc said:
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?

If it helps you, sure.
 
  • #12
L²Cc said:
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?

Only if afterwards you plug the (k+x)'s back in, so you can see what your new thing looks like.

And I disagree, the problem you posted does deal with factorials.


Just to confirm, you did figure out how the first problem became (n+2)! - 1 right?
 

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