SOLUTION: the directions of the textbook say write an expression for the apparent nth term of the sequence.( assume that n begiins with 1) and for 43. it has the numbers 0, 3, 8, 15, 24 h

Algebra ->  Algebra  -> Sequences-and-series -> SOLUTION: the directions of the textbook say write an expression for the apparent nth term of the sequence.( assume that n begiins with 1) and for 43. it has the numbers 0, 3, 8, 15, 24 h      Log On

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Question 107351This question is from textbook precalculus
: the directions of the textbook say write an expression for the apparent nth term of the sequence.( assume that n begiins with 1)
and for 43. it has the numbers 0, 3, 8, 15, 24
how would you do thatThis question is from textbook precalculus

Answer by Fombitz(13823) About Me  (Show Source):
You can put this solution on YOUR website!
n=1, 0
n=2, 3
n=3, 8
n=4, 15
n=5, 24
If I add 1 to each output I have a perfect square.
The function is then is a perfect square minus 1.
f%28n%29=n%5E2-1
The other pattern is that the difference between two output numbers is a consecutive odd number.
f%28n%2B1%29-f%28n%29=%28%28n%2B1%29%5E2-1%29-%28n%5E2-1%29%29
f%28n%2B1%29-f%28n%29=%28%28n%5E2%2B2n%2B1-1%29-n%5E2%2B1%29
f%28n%2B1%29-f%28n%29=+2n%2B1
That's just interesting more than anything else.
Now back to the problem.
The expression for the nth term is then, as above,
f%28n%29=n%5E2-1
and the value for the 43rd term is then,
f%28n%29=n%5E2-1
f%2843%29=43%5E2-1
f%2843%29=1848