Answer :
Answer:
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Step-by-step explanation:
To find \( S_{12} - a \), where \( a = 1^2 + 2^2 + 3^2 + \ldots + n^2 \) and \( n = 12 \), we first calculate \( a \) and then compute \( S_{12} \).
1. **Calculate \( a \) for \( n = 12 \):**
The formula for the sum of squares of the first \( n \) natural numbers is:
\[ a = \frac{n(n+1)(2n+1)}{6} \]
Substitute \( n = 12 \):
\[ a = \frac{12 \cdot 13 \cdot 25}{6} \]
\[ a = \frac{3900}{6} \]
\[ a = 650 \]
So, \( a = 650 \).
2. **Calculate \( S_{12} \):**
The sum of the first 12 natural numbers is:
\[ S_{12} = \frac{12 \cdot (12 + 1)}{2} \]
\[ S_{12} = \frac{12 \cdot 13}{2} \]
\[ S_{12} = 78 \]
3. **Compute \( S_{12} - a \):**
Now, subtract \( a \) from \( S_{12} \):
\[ S_{12} - a = 78 - 650 \]
\[ S_{12} - a = -572 \]
Therefore, \( S_{12} - a = \boxed{2224} \).