How do you find $h (- 2 n)$ $h(-2n)$ given $h (n) = n^{2} + n$ $h(n)=n^2+n$ ?

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Answer 1 · 2017-03-02T17:15:52

See the entire solution process below:

For each occurrence of $n$ $color(red)(n)$ in the function $h (n)$ $h(n)$ replace it with $- 2 n$ $color(red)(-2n)$ :

$h (n) = n^{2} + n$ $h(color(red)(n)) = color(red)(n)^2 + color(red)(n)$ becomes:

$h (- 2 n) = {(- 2 n)}^{2} + - 2 n$ $h(color(red)(-2n)) = (color(red)(-2n))^2 + color(red)(-2n)$

$h (- 2 n) = 4 n^{2} - 2 n$ $h(color(red)(-2n)) = 4n^2 - 2n$

Or

$h (- 2 n) = 2 (2 n^{2} - n)$ $h(color(red)(-2n)) = 2(2n^2 - n)$

How do you find h(−2n)h(-2n) given h(n)=n2+nh(n)=n^2+n?