f(n) = 4f(n−1) −4f(n−2) + n^2 and f(0)=2 , f(1)=5


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Question Number 129403 by bemath last updated on 15/Jan/21

$f (n) = 4 f (n - 1) - 4 f (n - 2) + n^{2}$ $and f (0) = 2, f (1) = 5$
Commented by talminator2856791 last updated on 15/Jan/21

$what is the question ?$
	Answered by mr W last updated on 15/Jan/21

	$l e t f_{n} = g_{n} + {A n}^{2} + B n + C$ $g_{n} + {A n}^{2} + B n + C = 4 g_{n - 1} - 4 g_{n - 2} + 4 A [{(n - 1)}^{2} - {(n - 2)}^{2}] + 4 B [(n - 1) - (n - 2)] + 4 (C - C) + n^{2}$ $g_{n} + {A n}^{2} + B n + C = 4 g_{n - 1} - 4 g_{n - 2} + 8 A n - 12 A + 4 B + n^{2}$ $\Rightarrow A = 1$ $\Rightarrow B = 8$ $\Rightarrow C = 20$ $g_{n} - 4 g_{n - 1} + 4 g_{n - 2} = 0$ $t^{2} - 4 t + 4 = 0$ ${(t - 2)}^{2} = 0$ $\Rightarrow t = 2$ $g_{n} = (α + β n) 2^{n}$ $\Rightarrow f (n) = (α + β n) 2^{n} + n^{2} + 8 n + 20$ $f (0) = α + 20 = 2 \Rightarrow α = - 18$ $f (1) = (- 18 + β) 2 + 1 + 8 + 20 = 5 \Rightarrow β = 6$ $\Rightarrow f (n) = 6 (n - 3) 2^{n} + n^{2} + 8 n + 20$
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