x


Question and Answers Forum

All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 214712 by efronzo1 last updated on 17/Dec/24

$\underset{―}{\underset{⏟}{x}}$
	Answered by mr W last updated on 17/Dec/24

	$r^{2} - 3 r + 2 = 0$ $(r - 1) (r - 2) = 0$ $r = 1, 2$ $\Rightarrow a_{n} = A \times 1^{n} + B \times 2^{n}$ $a_{0} = A + B = 2$ $a_{1} = A + 2 B = 5$ $\Rightarrow B = 3, A = - 1$ $\Rightarrow a_{n} = 3 \times 2^{n} - 1 ✓$
	Answered by mr W last updated on 17/Dec/24

	$a l t e r n a t i v e a p p r o a c h$ $a_{n + 2} - a_{n + 1} = 2 (a_{n + 1} - a_{n})$ $s a y b_{n} = a_{n + 1} - a_{n}$ $b_{n + 1} = 2 b_{n} \leftarrow G . P .$ $\Rightarrow b_{n} = 2^{n} \times b_{0} = 2^{n} (a_{1} - a_{0}) = 3 \times 2^{n}$ $a_{n + 1} - a_{n} = 3 \times 2^{n}$ $a_{n} - a_{n - 1} = 3 \times 2^{n - 1}$ $. . . . . .$ $a_{1} - a_{0} = 3 \times 2^{0}$ $a_{n + 1} - a_{0} = 3 (1 + 2 + . . . + 2^{n}) = 3 \times \frac{2^{n + 1} - 1}{2 - 1}$ $a_{n + 1} = 3 \times \frac{2^{n + 1} - 1}{2 - 1} + 2 = 3 \times 2^{n + 1} - 1$ $\Rightarrow a_{n} = 3 \times 2^{n} - 1 ✓$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum