Proving 2^n via Induction: 1+n!/1!+n(n-1)!/2!+...”

  • Context: Undergrad 
  • Thread starter Thread starter theperthvan
  • Start date Start date
  • Tags Tags
    Induction
Click For Summary
SUMMARY

The discussion focuses on proving the equation 1 + n/1! + n(n-1)/2! + n(n-1)(n-2)/3! + ... = 2^n through mathematical induction. This expression represents the sum of the binomial coefficients, which corresponds to the power set of a set with n elements, confirming that the number of subsets is 2^n. The binomial expansion of (1 + 1)^n is highlighted as a crucial hint for completing the proof.

PREREQUISITES
  • Understanding of mathematical induction
  • Familiarity with binomial coefficients
  • Knowledge of factorial notation and calculations
  • Basic concepts of set theory and power sets
NEXT STEPS
  • Study the principles of mathematical induction in depth
  • Explore the binomial theorem and its applications
  • Review the properties and calculations of factorials
  • Investigate the relationship between set theory and combinatorics
USEFUL FOR

Mathematicians, educators, students studying combinatorics, and anyone interested in proofs involving induction and binomial expansions.

theperthvan
Messages
182
Reaction score
0
How can I show that

1+\frac{n}{1!}+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!}+...= 2^{n}

This comes from proving that the power set of a set with n elements is 2^{n}.

I got so far that nCn+nC(n-1)+ ... = what I have above. Now for the induction...
Cheers,
 
Physics news on Phys.org
Hint: What's the binomial expansion of (1 + 1)^n?
 
of course, thanks
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K