If A=3
B=11
C=107
D= ?
Then what is E?
Im so confused because D isn't filled in....
You understand, do you not, that "given" just a list of numbers, there is no reason for the list to have any "rule" at all?
There is absolutely NO reason this list could not have any numbers whatsoever for D and E! Unless there is some other information this is really just an exercise in
guessing what the person who made up the list had in mind- an exercise in mind reading!
If we were to take these numbers to be the output of some function at n= 0, 1, 2, 3, and 4, the
simplest "rule" would be linear but if y= an+ b, we would have to have y(0)= a(0)+ b= b= 3 and y(1)= a(1)+ b= a+ 3= 11 so that a= 8. But then y(2)= 8(2)+ 3= 16+ 3= 19, not 107.
The simplest function that will give
three arbitrary values, here y(0)= 3, y(1)= 11, y(2)= 107, is a quadratic:
y(n)= an^2+ bn+ c. We must have y(0)= c= 3, y(1)= a+ b+ c= 11, y(2)= 4a+ 2b+ c= 107. Since c= 3, a+ b+ c= a+ b+ 3= 11 and 4a+ 2b+ 3= 107 so a+ b= 8 and 4a+ 2b= 104. If we multiply a+ b= 8 by 2, 2a+ 2b= 16, and subtract that from 4a+ 2b= 104, we eliminate b: 2a= 88 so that a= 44. Then a+ b= 44+ b= 8 so b= 8- 44= -36. The "rule" y(n)= 44n^2- 36n+ 3 gives y(0)= 3, y(1)= 44- 36+ 3= 8+ 3= 11, and y(2)= 44(4)- 36(2)+ 3= 176- 72+ 3= 104+3= 107. Now calculate y(3) and y(4) to find D and E.
HOWEVER, this is
one possible rule to give these values for A, B, and C. There is no way of knowing what the person who made up this problem had in mind.
PLEASE I have no idea how to even begin to figure this out!!!!!!!
please help me!
