Question

Which polynomial sequence shows that the terms of the second differences are constant

Answer 1

The property of constant second differences is characteristic of quadratic polynomials. A quadratic polynomial can be expressed in the general form:

\[ f(n) = an^2 + bn + c \]

where \( a \), \( b \), and \( c \) are constants, and \( n \) is the term number (1, 2, 3, ...).

To show that the second differences are constant, we can compute the first differences and then the second differences.

1. **First Differences**:
The first difference of a sequence \( f(n) \) is given by:

\[ \Delta f(n) = f(n+1) - f(n) \]

For the quadratic polynomial, we compute:

\[
f(n+1) = a(n+1)^2 + b(n+1) + c = a(n^2 + 2n + 1) + b(n + 1) + c = an^2 + (2a + b)n + (a + b + c)
\]

Hence,

\[
\Delta f(n) = f(n+1) - f(n) = \left[ an^2 + (2a + b)n + (a + b + c) \right] - \left[ an^2 + bn + c \right]
\]

This simplifies to:

\[
\Delta f(n) = (2a)n + (a + b)
\]

2. **Second Differences**:
Now we calculate the second difference, which is given by:

\[ \Delta^2 f(n) = \Delta f(n+1) - \Delta f(n) \]

Calculating \( \Delta f(n+1) \):

\[
\Delta f(n+1) = (2a)(n+1) + (a + b) = (2a)n + (2a) + (a + b)
\]

Now, calculate \( \Delta^2 f(n) \):

\[
\Delta^2 f(n) = \left[(2a)n + (2a + a + b)\right] - \left[(2a)n + (a + b)\right]
\]
\[
= (2a) - (0) = 2a
\]

The second differences \( \Delta^2 f(n) \) is constant and equal to \( 2a \). This confirms that the terms of the second differences are constant for any quadratic polynomial, implying that any quadratic polynomial will yield constant second differences in its sequence of values.