To find the third term in the expansion of \((a + 5)^5\) using Pascal's Triangle, we start by identifying the coefficients from the triangle for \(n = 5\).
The coefficients from Pascal's Triangle for \(n = 5\) are: \[ 1, 5, 10, 10, 5, 1 \]
In the binomial expansion of \((x + y)^n\), the general term can be represented as: \[ T_k = \binom{n}{k} x^{n-k} y^k \] where \(T_k\) is the \((k+1)\)-th term in the expansion.
In our case, \(n = 5\), \(x = a\), and \(y = 5\). We need to find the third term, which corresponds to \(k = 2\) (since \(k\) starts from 0).
Using the formula: \[ T_2 = \binom{5}{2} a^{5-2} (5)^2 \]
First, we calculate the binomial coefficient: \[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
Next, we evaluate \(a^{5-2}\): \[ a^{5-2} = a^3 \]
We also calculate \(5^2\): \[ 5^2 = 25 \]
Now we combine all parts to find the third term: \[ T_2 = 10 \cdot a^3 \cdot 25 = 250 a^3 \]
Thus, the third term in the expansion of \((a + 5)^5\) is: \[ \boxed{250a^3} \]
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