How do you calculate the nth term for 4, 10, 18, 28?

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To calculate the nth term for 4, 10, 18, 28, use the formula \( n^2 + 3n \).

To find the nth term of a sequence, we first need to identify the pattern or rule that generates the sequence. For the sequence 4, 10, 18, 28, we start by examining the differences between consecutive terms. The differences are 6, 8, and 10. These differences are not constant, so we look at the second differences, which are 2 and 2. Since the second differences are constant, this indicates a quadratic sequence.

A quadratic sequence can be expressed in the form \( an^2 + bn + c \). To find the coefficients \( a \), \( b \), and \( c \), we use the first few terms of the sequence.

1. For \( n = 1 \): \( a(1)^2 + b(1) + c = 4 \)
2. For \( n = 2 \): \( a(2)^2 + b(2) + c = 10 \)
3. For \( n = 3 \): \( a(3)^2 + b(3) + c = 18 \)

This gives us the system of equations:
1. \( a + b + c = 4 \)
2. \( 4a + 2b + c = 10 \)
3. \( 9a + 3b + c = 18 \)

By solving these equations, we find:
- Subtract the first equation from the second: \( 3a + b = 6 \)
- Subtract the second equation from the third: \( 5a + b = 8 \)

Now, subtract the first new equation from the second new equation:
- \( 2a = 2 \) so \( a = 1 \)

Substitute \( a = 1 \) back into \( 3a + b = 6 \):
- \( 3(1) + b = 6 \) so \( b = 3 \)

Finally, substitute \( a = 1 \) and \( b = 3 \) back into \( a + b + c = 4 \):
- \( 1 + 3 + c = 4 \) so \( c = 0 \)

Thus, the nth term formula is \( n