Binomial Theorem: Evaluating Complex Combinations

ritwik06
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Homework Statement



Evaluate
\sum^{m}_{r=0} ^{ n + r }C_{n}

I can handle things when the lower thing in the combination part is changing, what shall I do with this one?
 
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Try writing out a few terms in the series and see if it helps.
 
I get this:
^{n}C_{n} + ^{n+1}C_{n} + ^{n+2}C_{n} + ... + ^{n+r}C_{n}<br />
^{n}C_{0} + ^{n+1}C_{1} + ^{n+2}C_{2} + ... + ^{n+r}C_{r}<br />

All I can do is this, now both the superscript and th subscript are increasing in A.P.
 
The thread is still unsolved...
 
the suggestion was that you actually look at a few specific examples.
If m= 1, you have
^nC_n+ ^{n+1}C_n= \frac{n!}{n!0!}+ \frac{(n+1)!}{n!1!}= 1+ n+ 1= n+ 2
If m= 2, you have
^nC_n+ ^{n+1}C_n+ ^{n+2}C_n= n+ 2+ \frac{(n+2)!}{n! 2!}= n+ 2+ (n+1)(n+2)/2= n+2+ \frac{1}{2}n^2+ \frac{3}{2}x+ 1= \frac{1}{2}n^2+ \frac{5}{2}n+ 3

Try a few more like that and see if anything comes to mind.
 
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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