Conisder the recurrance relation $a_{k} = - 8 a_{k - 1} - 15 a_{k - 2}$ with initial condition $a_{0} = 0 a n d a_{1} = 2$ . which of the following is an explicit solution to this recurrance relation ?

k (- 3)^{k} - (- 5)^{k}

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(- 3)^{k} - (- 5)^{k}

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k (- 3)^{k} - k (- 5)^{k}

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(- 5)^{k} - (- 3)^{k}

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Solution

The correct option is B $(- 3)^{k} - (- 5)^{k}$
$a_{k} = - 8 a_{1 - 1} - 15 a_{k - 2}$
$C o n s i d e r k - 2 = 1, k - 1 = n a n d k = n^{2}$
$n^{2} = - 8 n - 15 [U s i n g c h a r a c t e r i s t i c e e q u a t i o n m e t h o d]$
$n^{2} + 8 n + 15 = 0$
$n^{2} + 5 n + 3 n + 15 = 0$
$n (n + 5) + 3 (n + 5) = 0$
$(n + 3) (n + 5) = 0$

$S o, n = - 3 a n d - 5$
$n_{1} = (- 3)^{1} C_{1} + (- 5)^{1} C_{2}$
$= (- 3)^{0} C_{1} + (- 5)^{0} C_{2} = 0$
$\Rightarrow C_{1} + C_{2} = 0 . . . (1)$
$a_{1} = (- 3)^{1} C_{1} + (- 5)^{1} C_{2} = 2$
$\Rightarrow - 3 C_{1} + - 5 C_{2} = 2 . . . (2)$

Solving equation (1) and (2), we get C $C_{1} = 1 a n d C_{2} = - 1$
$S o, a_{4} = (- 3)^{k} (- 5)^{1}$

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