Proof of Binomial Identity: Proving SUM(nCk)*2^k=(3^n+(-1)^n)/2

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SUMMARY

The forum discussion centers on proving the binomial identity SUM(nCk)*2^k=(3^n+(-1)^n)/2 for all positive integers n, with the condition that k is even. The Binomial Theorem is utilized, specifically applying it to the expressions (2+1)^n and (1-2)^n to derive the necessary components of the proof. The discussion emphasizes the importance of manipulating these binomial expansions to achieve the desired equality.

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  • Understanding of the Binomial Theorem
  • Familiarity with binomial coefficients (nCk)
  • Basic knowledge of algebraic manipulation
  • Experience with summation notation
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  • Explore advanced applications of the Binomial Theorem in combinatorial proofs
  • Study the properties of binomial coefficients and their generating functions
  • Investigate the implications of alternating series in binomial identities
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Mathematicians, students studying combinatorics, and anyone interested in advanced algebraic proofs will benefit from this discussion.

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Homework Statement


Prove that for all positive integers n, the equality holds:
SUM(nCk)*2^k=(3^n+(-1)^n)/2
Note: The sum goes from k=0 to n. AND k has to be even.


Homework Equations


Binomial Theorem


The Attempt at a Solution


I know that if we use the binomial theorem for x=2 and y=1, we would get
(2+1)^n=SUM(nCk)*2^k , with no restriction for k being even. I wonder how to proceed.
Thanks.
 
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Hint: Your idea looks like a good start. You have done (writing it in the other order) the expansion of (1 + 2)n.

Look at the same thing for (1 - 2)n and see if that gives you any ideas.
 

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