To fill in the blanks for the identity \((n - k + 2)^2\), we first need to expand the left-hand side.
Starting with the expression:
\[ (n - k + 2)^2 \]
We can expand it using the formula for the square of a binomial \((a + b + c)^2\):
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc \]
Here, we let:
- \(a = n\)
- \(b = -k\)
- \(c = 2\)
Now expanding this gives us:
\[ (n - k + 2)^2 = n^2 + (-k)^2 + 2^2 + 2(n)(-k) + 2(n)(2) + 2(-k)(2) \]
Calculating each term:
- \(n^2\) remains as \(n^2\)
- \((-k)^2 = k^2\)
- \(2^2 = 4\)
- \(2(n)(-k) = -2nk\)
- \(2(n)(2) = 4n\)
- \(2(-k)(2) = -4k\)
So when we combine all these terms, we have:
\[ (n - k + 2)^2 = n^2 + k^2 + 4 + (-2nk) + (4n) + (-4k) \]
Rearranging this, we get:
\[ (n - k + 2)^2 = n^2 + k^2 + 4 + 4n - 4k - 2nk \]
Now we can fill in the blanks:
- The coefficient of \(n^2\) is \(1\).
- The coefficient of \(k^2\) is \(1\).
- The coefficient of \(n\) is \(4\).
- The coefficient of \(k\) is \(-4\).
- The coefficient of \(nk\) is \(-2\).
- The constant term is \(4\).
So, we get:
\[ (n - k + 2)^2 = 1 \cdot n^2 + 1 \cdot k^2 + 4 \cdot n - 4 \cdot k - 2 \cdot nk + 4 \]
Thus the filled blanks are:
\[ 1, 1, 4, -4, -2, 4 \]
So, the final answer is:
\[ (n - k + 2)^2 = 1n^2 + 1k^2 + 4n - 4k - 2nk + 4 \]
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