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if-a-1-3-and-a-n-1-3a-n-6n-2-12n-2-find-a-n-in-terms-of-n-

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Question Number 77799 by mr W last updated on 10/Jan/20
if a_1 =3 and a_(n+1) =3a_n +6n^2 −12n+2 find a_n in terms of n.
ifa1=3andan+1=3an+6n2−12n+2findanintermsofn.
Answered by mr W last updated on 12/Jan/20
let A_n =3^n a+bn^2 +cn+d A_(n+1) =3^(n+1) a+b(n+1)^2 +c(n+1)+d =3^(n+1) a+3bn^2 +3cn+3d+6n^2 −12n+2 b(2n^2 −2n−1)+c(2n−1)+2d+6n^2 −12n+2=0 2(b+3)n^2 −2(b−c+6)n−b−c+2d+2=0 b+3=0 ⇒b=−3 b−c+6=0 ⇒−3−c+6=0 ⇒c=3 −b−c+2d+2=0 ⇒3−3+2d+2=0 ⇒d=−1 A_n =3^n a−3n^2 +3n−1 A_1 =3a−3+3−1=3 ⇒a=(4/3) ⇒A_n =4×3^(n−1) −3n^2 +3n−1 check: A_(n+1) =4×3^n −3(n+1)^2 +3(n+1)−1 A_(n+1) =4×3^n −3n^2 −3n−1 3A_n +6n^2 −12n+2=4×3^n −9n^2 +9n−3+6n^2 −12n+2 3A_n +6n^2 −12n+2=4×3^n −3n^2 −3n−1 ⇒A_(n+1) =3A_n +6n^2 −12n+2 ⇒ok
letAn=3na+bn2+cn+dAn+1=3n+1a+b(n+1)2+c(n+1)+d=3n+1a+3bn2+3cn+3d+6n2−12n+2b(2n2−2n−1)+c(2n−1)+2d+6n2−12n+2=02(b+3)n2−2(b−c+6)n−b−c+2d+2=0b+3=0⇒b=−3b−c+6=0⇒−3−c+6=0⇒c=3−b−c+2d+2=0⇒3−3+2d+2=0⇒d=−1An=3na−3n2+3n−1A1=3a−3+3−1=3⇒a=43⇒An=4×3n−1−3n2+3n−1check:An+1=4×3n−3(n+1)2+3(n+1)−1An+1=4×3n−3n2−3n−13An+6n2−12n+2=4×3n−9n2+9n−3+6n2−12n+23An+6n2−12n+2=4×3n−3n2−3n−1⇒An+1=3An+6n2−12n+2⇒ok

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