Proving Two Equations Involving nCr and Powers of 2

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SUMMARY

The discussion centers on proving two mathematical equations involving combinations (nCr) and powers of 2. The first equation states that the sum from r=0 to n of [((-1)^r) * (nCr)] equals 0, which can be derived from the expansion of (1-1)^n. The second equation asserts that the sum from r=0 to n of [nCr] equals 2^n, derived from the expansion of (1+1)^n. Both proofs utilize the binomial theorem as a foundational concept.

PREREQUISITES
  • Understanding of binomial coefficients (nCr)
  • Familiarity with the binomial theorem
  • Knowledge of alternating series
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the binomial theorem in detail, focusing on its applications
  • Learn about the properties of alternating series
  • Explore mathematical induction techniques for proofs
  • Investigate the implications of (1-1)^n and (1+1)^n in combinatorial proofs
USEFUL FOR

Students of mathematics, particularly those studying combinatorics, algebra, or preparing for advanced topics in proofs and series. This discussion is beneficial for anyone looking to deepen their understanding of binomial coefficients and their applications in proofs.

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Homework Statement


Prove:
The sum of r=0 to n [((-1)^r) * (nCr)] = 0.

Prove:
The sum of r=0 to n [nCr] = 2^n


Homework Equations


It says to consider (1-1)^n and (1+1)^n , but I have no idea what this is even relating to honestly.


The Attempt at a Solution


I need help getting started.
 
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You do know that
[tex](a+ b)^n= \sum_{r=0}^n \left(\begin{array}{c}n \\ r\end{array}\right)a^{n-r}b^r[/tex]
don't you?
 


Yes I do, but I don't know how to apply it. I realize the (1-1)^n is the considered piece for the alternating series. But do I just Induction with it?
 

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