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.
