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преобразование Лапласа 2cos(x)

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-й член a_n+1=−3a_n, a₁=8

Solution

a_n=8(−3)ⁿ⁻¹

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a₁=8, a_n+1=−3a_n

Найдите тип последовательности для a₁=8, a_n+1=−3a_n: геометрическая последовательность: r=−3

N-й член геометрической прогрессии имеет вид a_n=a₁·rⁿ⁻¹

a₁=8, r=−3

a_n=8(−3)ⁿ⁻¹

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тип последовательности a_n+1=−3a_n, a₁=8

первый член a_n+1=−3a_n, a₁=8

следующий термин a_n+1=−3a_n, a₁=8

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

преобразование Лапласа 2cos(x)

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