Trying to prove this equality involving a summation of a binomial coefficient.

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.

1-1.png


2-1.png
 
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hi jdinatale! :smile:

(i haven't looked at your induction proof , but …)

why not just multiply the LHS by n+1 ? :wink:
 
tiny-tim said:
hi jdinatale! :smile:

(i haven't looked at your induction proof , but …)

why not just multiply the LHS by n+1 ? :wink:

Ok, I tried that and I eventually could not go any further. Any ideas on what's going wrong?

3-1.png
 
hi jdinatale! :smile:

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? :wink:​
 
tiny-tim said:
hi jdinatale! :smile:

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? :wink:​

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.
 
easy! :smile:

the clue was in the question …

the RHS said 2n+1,

which i know is ∑ n+1Cr :wink:​
 

Thread 'Finding the nth roots of a complex number'

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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