a_(n+2) −3a_(n+1) +2a_n =n+3^n please find a


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Question Number 176708 by Matica last updated on 25/Sep/22

$a_{n + 2} - 3 a_{n + 1} + 2 a_{n} = n + 3^{n}$ $p l e a s e f i n d a_{n}$
	Answered by mr W last updated on 26/Sep/22

	$(a_{n + 2} - a_{n + 1}) - 2 (a_{n + 1} - a_{n}) = n + 3^{n}$ $l e t b_{n} = a_{n + 1} - a_{n}$ $b_{n + 1} - 2 b_{n} = n + 3^{n}$ $l e t b_{n} = c_{n} + p + q n + 3^{n} k$ $b_{n + 1} = c_{n + 1} + p + q (n + 1) + 3^{n + 1} k$ $c_{n + 1} - 2 c_{n} + p + q (n + 1) - 2 p - 2 q n + 3^{n + 1} k - 3^{n} k = n + 3^{n}$ $c_{n + 1} - 2 c_{n} + q - p - q n + 2 k 3^{n} = n + 3^{n}$ $\Rightarrow q = - 1$ $\Rightarrow q - p = 0 \Rightarrow p = - 1$ $\Rightarrow 2 k = 1 \Rightarrow k = \frac{1}{2}$ $\Rightarrow b_{n} = c_{n} - 1 - n + \frac{3^{n}}{2}$ $\Rightarrow c_{n + 1} - 2 c_{n} = 0$ $\Rightarrow c_{n} = 2 c_{n - 1} = 2^{2} c_{n - 2} = . . . = 2^{n} c_{0}$ $c_{0} = b_{0} + 1 + 0 + \frac{3^{0}}{2} = b_{0} + \frac{3}{2} = a_{1} - a_{0} + \frac{3}{2}$ $\Rightarrow b_{n} = a_{n + 1} - a_{n} = 2^{n} c_{0} - 1 - n + \frac{3^{n}}{2}$ $\underset{n = 1}{\sum^{n}} (a_{n + 1} - a_{n}) = \underset{n = 1}{\sum^{n}} (2^{n} c_{0} - 1 - n + \frac{3^{n}}{2})$ $a_{n + 1} - a_{1} = \frac{2 c_{0} (2^{n} - 1)}{2 - 1} - n - \frac{n (n + 1)}{2} + \frac{3 (3^{n} - 1)}{2 (3 - 1)}$ $a_{n + 1} = a_{1} + 2 c_{0} (2^{n} - 1) - \frac{n (n + 3)}{2} + \frac{3^{n + 1} - 3}{4}$ $\Rightarrow a_{n} = a_{1} + (a_{1} - a_{0} + \frac{3}{2}) (2^{n} - 2) - \frac{(n - 1) (n + 2)}{2} + \frac{3^{n} - 3}{4}$
	Commented by Matica last updated on 26/Sep/22

	$T h a n k y o u v e r y m u c h$
	Commented by Tawa11 last updated on 25/Sep/22

	$Great sir$
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