Calculating the Value of Combinations with Induction

Click For Summary

Homework Help Overview

The discussion revolves around calculating the value of combinations defined as (n k) = n!/k!(n-k)! and evaluating the sum of these combinations from k=0 to n. Participants are exploring methods to prove that this sum equals 2^n, considering various mathematical approaches.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to evaluate the combinations for specific values of n and questions whether they are on the right track. Some participants suggest using the Binomial Theorem to simplify the problem, while others inquire about alternative methods to reach the result without relying on the theorem.

Discussion Status

Participants are actively discussing different approaches, including the use of induction and the Binomial Theorem. There is an acknowledgment of the potential equivalence of the sum to 2^n, but no consensus has been reached on the methods to prove it.

Contextual Notes

Some participants express uncertainty about the necessity of calculus in solving the problem and are exploring the implications of using different mathematical tools to arrive at the solution.

SMA_01
Messages
215
Reaction score
0

Homework Statement



Let n, k be in the set of all positive integers including zero. Define (n k)=n!/k!(n-k)!
Determine the value of [itex]\sum[/itex] (n k) from k=0 to n. Determine the value of [itex]\sum[/itex] (n k) from k=0 to n.

The Attempt at a Solution



I tried evaluating for some n value:

(n 0)= 1

(n 1)= n

(n 2)= (1/2!)n(n-1)

(n 3)= (1/3!)n(n-1)(n-2)So, generally= 1+n+...+[(1/k!)(n)*...*(n-(n-k))]+1

Am I on the right track? How can I determine the value? Any help is appreciated.
 
Physics news on Phys.org
I don't think that this problem requires use of Calculus, it can be simply done using the Binomial Theorem.

What is (x+a)n? Write that and try to find the values of x and a by comparing the formula with the series you got i.e. with 1+n+...+[(1/k!)(n)*...*(n-(n-k))].

http://en.wikipedia.org/wiki/Binomial_theorem
 
Pranav-Arora: Thanks. I'm going to have to prove it using induction. Is there anyway other than the binomial theorem to reach the result? I know it will equal 2^n, but I was wondering if you can compute without using the binomial theorem...?
 
SMA_01 said:
Pranav-Arora: Thanks. I'm going to have to prove it using induction. Is there anyway other than the binomial theorem to reach the result? I know it will equal 2^n, but I was wondering if you can compute without using the binomial theorem...?

I am not sure if i know of any other method, you should better wait for others members to look at this thread. :)
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 22 ·
Replies
22
Views
2K
Replies
11
Views
2K
Replies
7
Views
4K