First, we can distribute the sum:
$\sum_{n=1}^{90}(3-2n) = \sum_{n=1}^{90}3 - \sum_{n=1}^{90}(2n)$
Now, we can separate the sums:
$\sum_{n=1}^{90}3 - \sum_{n=1}^{90}(2n) = 3\sum_{n=1}^{90}1 - 2\sum_{n=1}^{90}n$
We know that the sum of the first $n$ integers is
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