Binomial Theorem: Evaluating Complex Combinations

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Homework Help Overview

The discussion revolves around evaluating a summation involving binomial coefficients, specifically the expression \(\sum^{m}_{r=0} {^{n + r}C_{n}}\). The original poster expresses uncertainty about handling the changing upper index in the combination.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants suggest writing out terms of the series to gain insight. The original poster attempts to express the series in terms of specific binomial coefficients, noting the patterns in the indices. Others encourage exploring specific examples to identify potential patterns or results.

Discussion Status

The discussion remains open with no consensus reached. Some participants have provided examples for the original poster to consider, which may help in understanding the problem better.

Contextual Notes

There is an indication that the original poster is familiar with combinations but is struggling with the specific structure of the summation. The examples provided by others suggest a focus on evaluating the expression for small values of \(m\).

ritwik06
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Homework Statement



Evaluate
[tex]\sum^{m}_{r=0} ^{ n + r }C_{n}[/tex]

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:
[tex]^{n}C_{n} + ^{n+1}C_{n} + ^{n+2}C_{n} + ... + ^{n+r}C_{n}[/tex]
[tex]^{n}C_{0} + ^{n+1}C_{1} + ^{n+2}C_{2} + ... + ^{n+r}C_{r}[/tex]

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
[tex]^nC_n+ ^{n+1}C_n= \frac{n!}{n!0!}+ \frac{(n+1)!}{n!1!}= 1+ n+ 1= n+ 2[/tex]
If m= 2, you have
[tex]^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[/tex]

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

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