General formula the binomial theorem:
$(x+y)^n=\sum\limits_{j=0}^n{n\choose j}x^{n-j}y^j$, but then what does this formula look like $(x – y)^n$ ?
More precisely, what her non-strict recording sequence looks like, and just as her strict recording using a sigma icon looks like.
I need this because I'm trying to understand the principle of adding two sequences with a binomial coefficient.
Best Answer
$x - y = x+ (-y)= x+(-1)y$.
So $(x-y)^n =(x+(-y))^n=\sum\limits_{j=0}^n{n\choose j}x^{n-j}(-y)^j=$
$\sum\limits_{j=0}^n(-1)^j{n\choose j}x^{n-j}y^j$
It is presumed to be understood that $(-1)^j$ is $-1$ if $j$ is odd and is $1$ if $j$ is even.
So in essence: $(x- y)^n$ is much the same as $(x+y)^2$ but ever odd term is subtracted rather than added (and every even term is added).
So $(x+y)^n + (x-y)^n =\sum\limits_{j=0}^n{n\choose j}x^{n-j}y^j+ \sum\limits_{j=0}^n(-1)^j{n\choose j}x^{n-j}y^j=$
$\sum\limits_{j=0}(1+(-1)^j){n\choose j}x^{n-j}y^j=$
$\sum\limits_{j=0}^n\begin{cases}2 & \text{if j is even}\\0 &\text{if j is odd}\end{cases}{n\choose j}x^{n-j}y^j=$
$\sum\limits_{j=0;j \text{ is even}}^n2{n\choose j}x^{n-j}y^j$
$=$ "twice the even terms"
....
And $(x+y)^n - (x-y)^n=\sum\limits_{j=0}^n{n\choose j}x^{n-j}y^j- \sum\limits_{j=0}^n(-1)^j{n\choose j}x^{n-j}y^j=$
$\sum\limits_{j=0}(1-(-1)^j){n\choose j}x^{n-j}y^j=$
$\sum\limits_{j=0}^n\begin{cases}0 & \text{if j is even}\\2 &\text{if j is odd}\end{cases}{n\choose j}x^{n-j}y^j=$
$\sum\limits_{j=0;j \text{ is odd}}^n2{n\choose j}x^{n-j}y^j$
$=$ "twice the odd terms"