%i created a fibonacci sequence but i want an expanded output of the polynmial that give the fibonacii sequence, %ie (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

clc;

clear;

close all;

n = input('n:');

%Create first two rows that are always constant

pt(1, 1) = 1;

pt(2, 1 : 2) = [1 1];

for row = 3 : n

% First element of every row

pt(row, 1) = 1;

% Every element is the addition of the two elements

% on top of it. That means the previous row.

for column = 2 : row-1

pt(row, column) = pt(row-1, column-1) + pt(row-1, column);

end

% Last element of every row

pt(row, row) = 1;

end

fprintf('Pascals triangle:\n');

pt(row, column) = pt(row-1, column-1) + pt(row-1, column)

fprintf('Expansion result:\n')

fprintf('(x+y)^%d =',n);

x=n;

y=0;

[Pallav Patel]

Pascals triangle*

Answer 1

[Raunak Gupta]

Hi,

As per my understanding above code will give a pascal triangle if you store the value in pt as mentioned in the question. If the size of pt is (n+1)x(n+1) the last row will give the coefficient of polynomial expansion of (x+y)^n . For formatting the result correctly you can define two vectors with x exponent and y exponent respectively as n:-1:0 and 0:1:n and can format using for loop and sprintf and strcat.

You can also use symbolic toolbox function expand for nicely displaying the expansion of (x+y)^n.

syms x y 
n = 5; 
p = (x + y)^n;  
expand(p)