if a_1 =3 and a_(n+1) =3a_n +6n^2 −12n+2 find a


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Question Number 77799 by mr W last updated on 10/Jan/20

$i f a_{1} = 3 a n d$ $a_{n + 1} = 3 a_{n} + 6 n^{2} - 12 n + 2$ $f i n d a_{n} i n t e r m s o f n .$
	Answered by mr W last updated on 12/Jan/20

	$l e t A_{n} = 3^{n} a + {b n}^{2} + c n + d$ $A_{n + 1} = 3^{n + 1} a + b {(n + 1)}^{2} + c (n + 1) + d$ $= 3^{n + 1} a + 3 {b n}^{2} + 3 c n + 3 d + 6 n^{2} - 12 n + 2$ $b (2 n^{2} - 2 n - 1) + c (2 n - 1) + 2 d + 6 n^{2} - 12 n + 2 = 0$ $2 (b + 3) n^{2} - 2 (b - c + 6) n - b - c + 2 d + 2 = 0$ $b + 3 = 0 \Rightarrow b = - 3$ $b - c + 6 = 0 \Rightarrow - 3 - c + 6 = 0 \Rightarrow c = 3$ $- b - c + 2 d + 2 = 0 \Rightarrow 3 - 3 + 2 d + 2 = 0 \Rightarrow d = - 1$ $A_{n} = 3^{n} a - 3 n^{2} + 3 n - 1$ $A_{1} = 3 a - 3 + 3 - 1 = 3$ $\Rightarrow a = \frac{4}{3}$ $\Rightarrow A_{n} = 4 \times 3^{n - 1} - 3 n^{2} + 3 n - 1$ $c h e c k :$ $A_{n + 1} = 4 \times 3^{n} - 3 {(n + 1)}^{2} + 3 (n + 1) - 1$ $A_{n + 1} = 4 \times 3^{n} - 3 n^{2} - 3 n - 1$ $3 A_{n} + 6 n^{2} - 12 n + 2 = 4 \times 3^{n} - 9 n^{2} + 9 n - 3 + 6 n^{2} - 12 n + 2$ $3 A_{n} + 6 n^{2} - 12 n + 2 = 4 \times 3^{n} - 3 n^{2} - 3 n - 1$ $\Rightarrow A_{n + 1} = 3 A_{n} + 6 n^{2} - 12 n + 2 \Rightarrow o k$
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