What are the first 6 terms of the sequence with the explicit formula
an=-3n2+4?
let n = 1 which gives you 3(1) + 4
Let n = 2 gives you 3(2)^2 + 4
Continue like this until you have 6 terms.
Got it, thanks!
don't forget the pesky "-" sign there
To find the first 6 terms of the sequence with the explicit formula an = -3n^2 + 4, you can substitute different values for n into the formula and evaluate the expression.
Let's start with n = 1:
a1 = -3(1)^2 + 4
a1 = -3(1) + 4
a1 = -3 + 4
a1 = 1
Next, let's find the second term by substituting n = 2:
a2 = -3(2)^2 + 4
a2 = -3(4) + 4
a2 = -12 + 4
a2 = -8
Now, let's calculate the third term using n = 3:
a3 = -3(3)^2 + 4
a3 = -3(9) + 4
a3 = -27 + 4
a3 = -23
Continue this process for n = 4, 5, and 6 to find the next terms of the sequence.
n = 4:
a4 = -3(4)^2 + 4
a4 = -3(16) + 4
a4 = -48 + 4
a4 = -44
n = 5:
a5 = -3(5)^2 + 4
a5 = -3(25) + 4
a5 = -75 + 4
a5 = -71
n = 6:
a6 = -3(6)^2 + 4
a6 = -3(36) + 4
a6 = -108 + 4
a6 = -104
Therefore, the first 6 terms of the sequence with the explicit formula an = -3n^2 + 4 are:
a1 = 1, a2 = -8, a3 = -23, a4 = -44, a5 = -71, a6 = -104.