how to find the sum terms in of binomial expansion
显示 更早的评论
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
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
Ahmed A. Selman
2013-4-12
I only added the command for nCr and it worked fine as:
...
p=...;% works with p=0, 1, 2, -1
q=...;% works with q=0, 1, 3, 0
...
nCr = nchoosek(15,r);
...
Ahmed A. Selman
2013-4-12
As a matter of fact it is also expandable to a loop over n and r, keeping in mind that there is a restriction of (n-r) >= 0. So, let's try this:
clc
clear
p=1;
q=3;
i=0;
nTerms=zeros(15,15);
rTerms=zeros(1,15);
for r = 2:15
i = i+1;
for n=r:15;
nCr = nchoosek(n,r);
nTerms(i,n) = nCr * p.^r .* q.^(n-r);
end
rTerms(r)=sum(nTerms(:,n));
end
SumOfAllTerms = sum(rTerms);% Sum with condition (n-r)>=0
n15_Sum=sum(nTerms(:,15));% Only at r=2:15 and n=15
also works, error-free.
a
2013-4-12
a
2013-4-12
Image Analyst
2013-4-12
Paste it into a new editor window and run it from there, not the command line. The command line is mostly for running single statements, not huge groups of statements.
a
2013-4-12
Image Analyst
2013-4-12
Look in the variable editor, or use fprintf(), or take the semicolon off the end of the lines.
a
2013-4-12
更多回答（1 个）
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
在 帮助中心 和 File Exchange 中查找有关 Lighting, Transparency, and Shading 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 
