## Algebra 2 (1st Edition)

$d(n)=32n-16$
From part (a), we have: $d(n)=D(n)-D(n-1)$ Thus, we have $d(n)=16n^2 -16(n-1)^2$ When we plug in $n=1,2,3,4$: This gives: $16, 48,80,112$ $d(n)=16n^2 -16(n^2-2n+1)$ Thus, $d(n)=32n-16$