Let a be the value of the function when n is 0. The number of the first row of differences is b, the first number of the next row is c, and so on. The formula for the nth number of the series is:
a + bn + cn(n-1)/2! + dn(n-l)(n-2)/3! and so on.
Find the formula for each pattern.
Here is the one I'm having trouble with:
6, 15, 24, 39, 60.
Please help if you can!
Thank you! :-)
To find the formula for the given pattern, we need to analyze the differences between consecutive terms.
Let's start by calculating the differences between terms:
15 - 6 = 9
24 - 15 = 9
39 - 24 = 15
60 - 39 = 21
Now, let's continue by calculating the differences between these first-order differences:
9 - 9 = 0
15 - 9 = 6
21 - 15 = 6
We observe that the second-order differences, after the initial zero, are all equal to 6. This constant difference in the second-order differences suggests that the pattern may be an exponential one.
To confirm this, we look for the relationship between the second-order differences and the power of n. Calculating the third-order differences:
6 - 0 = 6
The third-order differences are constant and equal to 6. This indicates that the pattern may be a quadratic one.
Now, let's construct the general form of the formula you provided:
a + bn + cn(n-1)/2! + dn(n-1)(n-2)/3! + ...
Since the second-order differences are constant and equal to 6, we can assume that c = 6.
Therefore, the formula for the given pattern is:
a + bn + 6n(n-1)/2!
To determine the values of a and b, we can substitute the first two terms of the sequence into the formula.
Given that the first term is 6, when n = 0:
6 = a + b(0) + 6(0)(-1)/2!
6 = a
Given that the second term is 15, when n = 1:
15 = a + b(1) + 6(1)(0)/2!
15 = a + b
Since we know that a = 6, we can substitute it into the equation:
15 = 6 + b
b = 9
Therefore, the final formula for the given pattern is:
6 + 9n + 6n(n-1)/2!