- #1
jdinatale
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I immediately thought of induction, so that is what I used, but I can't seem to make any progress past a certain point.
Trying to prove this equality involving a summation of a binomial coefficient.
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- Binomial Coefficient Summation
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Homework Help Overview
The discussion revolves around proving an equality that involves a summation of binomial coefficients. The original poster attempts to use mathematical induction but encounters difficulties in progressing through the proof.
Discussion Character
- Exploratory, Mathematical reasoning, Assumption checking
Approaches and Questions Raised
- Participants discuss the use of induction and alternative approaches, such as manipulating the left-hand side of the equation. Questions arise regarding the steps taken in the proof and the reasoning behind certain transformations of the summation.
Discussion Status
The discussion is active, with participants offering guidance and alternative perspectives. Some have provided insights that appear to help clarify the problem, while others are exploring different interpretations of the steps involved in the proof.
Contextual Notes
There are indications of missing information or assumptions that may affect the proof, particularly regarding the manipulation of binomial coefficients and the interpretation of the summation limits.
- #2
tiny-tim
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hi jdinatale! 
(i haven't looked at your induction proof , but …)
why not just multiply the LHS by n+1 ?
(i haven't looked at your induction proof , but …)
why not just multiply the LHS by n+1 ?
- #3
jdinatale
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tiny-tim said:hi jdinatale!
(i haven't looked at your induction proof , but …)
why not just multiply the LHS by n+1 ?
Ok, I tried that and I eventually could not go any further. Any ideas on what's going wrong?
- #4
tiny-tim
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hi jdinatale!
in the third line you have ∑k=0…j j+1Ck+1
put m = k+1, that's ∑m=1…j+1 j+1Cm …
in the third line you have ∑k=0…j j+1Ck+1
put m = k+1, that's ∑m=1…j+1 j+1Cm …
what is that?
- #5
jdinatale
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tiny-tim said:hi jdinatale!
in the third line you have ∑k=0…j j+1Ck+1
put m = k+1, that's ∑m=1…j+1 j+1Cm …
what is that?
Brilliant! I've solved the problem now, thank you so much. But please tell me, how did you possibly know to do that? That was not obvious to me at all, and I'm not sure how you would just know to do that.
- #6
tiny-tim
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easy!
the clue was in the question …
the clue was in the question …
the RHS said 2n+1,
which i know is ∑ n+1Cr
which i know is ∑ n+1Cr
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