Question

Is binomial Theorem JUST Pascal's triangle? Or is there another equation...?

Answer 1

Pascal's triangle is a visual representation of the coefficients involved in a binomial expansion. The $n$th row of the triangle gives the coefficients of the terms in the expansion of $(a+b)^{n-1}$.

The triangle itself looks like

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\end{matrix}$

and so on, while the expansions for $n=1,2,3,4$ are

$(a+b)^{1-1}=(a+b)^0=1$
$(a+b)^{2-1}=(a+b)^1=1a+1b$
$(a+b)^{3-1}=(a+b)^2=1a^2+2ab+1b^2$
$(a+b)^{4-1}=(a+b)^3=1a^3+3a^2b+3ab^2+1b^3$

and so on.

The binomial theorem says that

$(a+b)^{n-1}=\displaystyle\sum_{k=0}^{n-1}\binom{n-1}ka^{n-1-k}b^k$
$(a+b)^{n-1}=\dbinom{n-1}0a^{n-1}b^0+\dbinom{n-1}1a^{n-2}b^1+\cdots+\dbinom{n-1}{n-2}a^1b^{n-2}+\dbinom{n-1}{n-1}a^0b^{n-1}$
$(a+b)^{n-1}=\dbinom{n-1}0a^{n-1}+\dbinom{n-1}1a^{n-2}b+\cdots+\dbinom{n-1}{n-2}ab^{n-2}+\dbinom{n-1}{n-1}b^{n-1}$

The numbers in the $n$th row of the triangle are just $\dbinom{n-1}k$, with $k=0,1,\ldots,n-2,n-1$.

So no, the binomial theorem and Pascal's triangle are not the same thing. Pascal's triangle is a way of organizing the pattern exhibited by the result of the binomial theorem.