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.