Albebra 2
posted by Lea .
Are these correct?
a1=(3x1)2 +4=13
a2=(3x2)2+440
a3=(3x3)2+4=85
a4=(3x4)2+4=148
a5=(3x4)2+4=229
a6=(3x4)2+4=328
The first six terms of the sequence are 13, 40, 85, 148, 229, 328
I keep coming up with these answers but there is suppose to be an error somewhere.

As you wrote it before,
3x^2+4 = 3(x^2) + 4
You have tabulated (3x)^2 + 4, which is very different
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