Prove this inequality with binomial

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SUMMARY

The discussion centers on proving the inequality \(\sum_{k=0}^n {3k\choose k} \ge \frac{5^n-1}{4}\). The user attempts to apply mathematical induction, starting with the base case \(n=0\), and assumes the inequality holds for \(n\). The next step involves evaluating \(\sum_{k=0}^{n+1} {3k\choose k}\) and recognizing the need to show that \({3(n+1)\choose (n+1)}\) contributes sufficiently to maintain the inequality. The conversation emphasizes the importance of simplifying the right-hand side and testing small values of \(n\) for insights.

PREREQUISITES
  • Understanding of binomial coefficients, specifically \({3k\choose k}\)
  • Familiarity with mathematical induction principles
  • Basic knowledge of inequalities and their proofs
  • Ability to manipulate algebraic expressions involving powers and factorials
NEXT STEPS
  • Study the properties of binomial coefficients, particularly \({3k\choose k}\)
  • Learn advanced techniques in mathematical induction
  • Explore simplification methods for algebraic inequalities
  • Investigate combinatorial proofs related to binomial inequalities
USEFUL FOR

Mathematics students, educators, and anyone interested in combinatorial proofs and inequalities will benefit from this discussion.

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Homework Statement


Prove that

[itex]\sum_{k=0}^n {3k\choose k}\ge \frac{5^n-1}{4}[/itex]

Homework Equations



[itex]{3k\choose k}= \frac{(3k)!}{k!(2k)!}[/itex]

The Attempt at a Solution



I tried using the induction principle, but...

Here my attempt:

For [itex]n=0[/itex] 1>0 ok

Suppose that is true for [itex]n[/itex], i.e.:

[itex]\sum_{k=0}^n {3k\choose k}\ge \frac{5^n-1}{4}[/itex]

Now:

[itex]\sum_{k=0}^{n+1} {3k\choose k}= \sum_{k=0}^{n} {3k\choose k}+ {3(n+1)\choose (n+1)}\ge \frac{5^n-1}{4}+{3(n+1)\choose (n+1)}[/itex]

But now I don't know what to do, maybe it is not the correct way to show this... I need your help
 
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Please, can someone help me? I think that it exists a different way to prove this inequality, but I don't know how to proceed :(
 
What you actually need to show is that C(3(n+1),(n+1))>=(5^(n+1)-1)/4-(5^n-1)/4. Do you see why? You can definitely simplify the right side a lot. Now look at both sides for small values of n. Can you see how to continue? Think about it. I haven't much beyond this. Help me!
 
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