how to find the sum terms in of binomial expansion

9 次查看(过去 30 天)
I need to find the sum of few terms in binomial expansion...more precisely i need to find the sum of this expression:
(nCr) * p^r * q^(n - r)
and limits for summation are from r = 2 to 15. and n=15
any help would be most welcome.

采纳的回答

Ahmed A. Selman
Ahmed A. Selman 2013-4-11
编辑:Ahmed A. Selman 2013-4-11
The general steps to find such a summation are:
- Start a loop over r,
- Calculate each term as a function of (r),
- In the loop, add the terms one by one to a unique matrix,
- After the loop is finished, sum over the added terms.
The code should be something as :
% ∑ (nCr) * p^r * q^(n - r)
clc; clear all
p =...;
q =...;
n = 15;
i = 0;
for r = 2:15
i = i+1;
nCr = ...;% calculate the coefficients here
Terms(i) = nCr * p.^r .* q.^(n-r);
end
SumOfTerms = sum(Terms)
The output is in (SumOfTerms), which should be a single value.
I assumed that (nCr) is not a constant, as I expect, it must be a function of (n and r). If it was a constant of (n and r), then define it outside the loop.
I also didn't understand the meaning of ( *| ) at the beginning and end of the formula in your question. If they mean some operation s ( complex conjugate or absolute) then add any of the commands at the very end of the code:
finalSum = conj(SumOfTerms);% for complex conjugate value.
or
finalSum = real(SumOfTerms);% for real value.
and in this case the output will be in (finalSum).
If you tried it and it didn't work, please let me know in which line it made an error, and the error message.
  11 个评论
a
a 2013-4-12
@ Ahmed A Selman:
thank You Mr. Ahmed A Selman for helping me out.
Regards,
a
a 2013-5-14
编辑:a 2013-5-14
@ Ahmed A. Selman
Sir, the code you provided works well...Can you please explain me how this loop is executing?

请先登录,再进行评论。

更多回答(1 个)

Roger Stafford
Roger Stafford 2013-4-11
With that range of r I would think it would be more efficient to compute
(q+p)^n-q^n-n*q^(n-1)*p

类别

Help CenterFile Exchange 中查找有关 Mathematics 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by