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Question Number 118849 by mr W last updated on 20/Oct/20

Commented by PRITHWISH SEN 2 last updated on 20/Oct/20

$(a_{n} + b_{n}) = 3^{n} - 1$
Commented by mr W last updated on 20/Oct/20

$f i n d a_{n} a n d b_{n} i n t e r m s o f n .$
	Answered by mr W last updated on 20/Oct/20

	$l e t$ $a_{n + 1} + {q b}_{n + 1} + r = p (a_{n} + {q b}_{n} + r)$ $a_{n + 1} + q (- 4 a_{n} - 3 b_{n} - 2) = {p a}_{n} + {p q b}_{n} + (p - 1) r$ $a_{n + 1} = (p + 4 q) a_{n} + (p + 3) {q b}_{n} + (p - 1) r + 2 q$ $\Rightarrow p + 4 q = 7$ $\Rightarrow (p + 3) q = 6$ $\Rightarrow (p - 1) r + 2 q = 4$ $\Rightarrow (7 - 4 q + 3) q = 6$ $\Rightarrow 2 q^{2} - 5 q + 3 = 0$ $\Rightarrow (2 q - 3) (q - 1) = 0$ $\Rightarrow q = 1 o r \frac{3}{2} (r e j e c t e d)$ $\Rightarrow p = 3 o r 1 (r e j e c t e d)$ $\Rightarrow r = 1$ $a_{n + 1} + b_{n + 1} + 1 = 3 (a_{n} + b_{n} + 1)$ $= 3^{n} (a_{1} + b_{1} + 1)$ $= 3^{n + 1}$ $\Rightarrow a_{n} + b_{n} = 3^{n} - 1$ $a_{n + 1} = 7 a_{n} + 6 (3^{n} - 1 - a_{n}) + 4 = a_{n} + 2 \times 3^{n + 1} - 2$ $\underset{k = 1}{\sum^{n}} a_{k + 1} = \underset{k = 1}{\sum^{n}} a_{k} + \underset{k = 1}{\sum^{n}} (2 \times 3^{k + 1} - 2)$ $a_{n + 1} - a_{1} = 2 \times 3^{2} \times \frac{3^{n} - 1}{2} - 2 n$ $a_{n + 1} = 3^{n + 2} - 2 n - 8$ $\Rightarrow a_{n} = 3^{n + 1} - 2 n - 6$ $\Rightarrow b_{n} = - 2 \times 3^{n} + 2 n + 5$
	Commented by PRITHWISH SEN 2 last updated on 21/Oct/20

	$excellent sir$
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