To find the expansion of \((a + 1)^3\) using Pascal's Triangle, we first identify the coefficients from the triangle for \(n = 3\).
The coefficients for \(n = 3\) are as follows:
\[ \begin{array}{cccc} & & 1 & \ & 1 & & 1 \ 1 & & & 1 \ \hline 1 & 3 & 3 & 1 \end{array} \]
Thus, the coefficients are \(1\), \(3\), \(3\), and \(1\).
The expansion of \((a + b)^n\) can be represented as:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For \((a + 1)^3\), we will have:
- For \(k = 0\): \(\binom{3}{0} a^{3} \cdot 1^0 = 1 \cdot a^3 = a^3\)
- For \(k = 1\): \(\binom{3}{1} a^{2} \cdot 1^1 = 3 \cdot a^2 = 3a^2\)
- For \(k = 2\): \(\binom{3}{2} a^{1} \cdot 1^2 = 3 \cdot a = 3a\)
- For \(k = 3\): \(\binom{3}{3} a^{0} \cdot 1^3 = 1 \cdot 1 = 1\)
Now, adding these terms together, we get:
\[ (a + 1)^3 = a^3 + 3a^2 + 3a + 1 \]
So, the expansion of \((a + 1)^3\) is:
\[ a^3 + 3a^2 + 3a + 1 \]