SOLUTION: The first 4 nth terms of 2n (squared) + 3n

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Question 342429: The first 4 nth terms of 2n (squared) + 3n
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first two.
You finish it.
n=1, 2n%5E2%2B3n=2%281%29%5E2%2B3%281%29=2%281%29%2B3=2%2B3=5
n=2, 2n%5E2%2B3n=2%282%29%5E2%2B3%282%29=2%284%29%2B6=8%2B6=14
Do it the same way.