Question

Why does the n+1 and (1-n) cancel out?


`4 {}^(n + 1) * 4 {}^(1 - n) * 4 {}^(2) \\ = 4 {}^(n + 1 + (1 - n) + 2) \\ = 4 {}^(4) \\ = 256`
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Answer 1

The rule being used is
`a^b*a^c = a^(b+c)`

You add the exponents when multiplying exponentials like this with the same base.

The n+1 adds with 1-n so that the n terms cancel out

n+(-n) = n-n = 0n = 0

The '1's dont cancel out though.