To find the second differences of the sequence generated by the polynomial \( f(n) = 2n^2 + 5 \), we need to first calculate the values of the polynomial for several integer values of \( n \). We will then find the first differences and subsequently the second differences.
Step 1: Calculate values for \( n \)
Let's calculate the values of \( f(n) \) for \( n = 1, 2, 3, 4, 5 \):
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For \( n = 1 \): \[ f(1) = 2(1^2) + 5 = 2(1) + 5 = 2 + 5 = 7 \]
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For \( n = 2 \): \[ f(2) = 2(2^2) + 5 = 2(4) + 5 = 8 + 5 = 13 \]
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For \( n = 3 \): \[ f(3) = 2(3^2) + 5 = 2(9) + 5 = 18 + 5 = 23 \]
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For \( n = 4 \): \[ f(4) = 2(4^2) + 5 = 2(16) + 5 = 32 + 5 = 37 \]
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For \( n = 5 \): \[ f(5) = 2(5^2) + 5 = 2(25) + 5 = 50 + 5 = 55 \]
Now we have the following values of \( f(n) \): \[ \begin{align*} f(1) & = 7 \ f(2) & = 13 \ f(3) & = 23 \ f(4) & = 37 \ f(5) & = 55 \ \end{align*} \]
Step 2: First Differences
Next, we calculate the first differences, which are the differences between consecutive values of \( f(n) \):
\[ \begin{align*} f(2) - f(1) & = 13 - 7 = 6 \ f(3) - f(2) & = 23 - 13 = 10 \ f(4) - f(3) & = 37 - 23 = 14 \ f(5) - f(4) & = 55 - 37 = 18 \ \end{align*} \]
So the first differences are: \[ \begin{align*} 6, 10, 14, 18 \end{align*} \]
Step 3: Second Differences
Now we calculate the second differences, which are the differences between consecutive first differences:
\[ \begin{align*} 10 - 6 & = 4 \ 14 - 10 & = 4 \ 18 - 14 & = 4 \ \end{align*} \]
So the second differences are: \[ \begin{align*} 4, 4, 4 \end{align*} \]
Conclusion
The second differences of the sequence generated by the polynomial \( f(n) = 2n^2 + 5 \) are constant and equal to \( 4 \). This indicates that the original sequence is quadratic, as expected for a polynomial of degree 2.
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