Recursive Sequence Growth

Question

A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the $n^{	ext{th}}$ day exceeds one million, then the lowest possible value of $n$ is

Accepted Answer

19. Let $x_n$ be the number of organisms on the $n^{	ext{th}}$ day. Given $x_1 = 2$ and $x_n = 2 x_{n-1} + 3$. Add 3 to both sides: $$x_n + 3 = 2 x_{n-1} + 6 = 2(x_{n-1} + 3)$$ Let $y_n = x_n + 3$. Then $y_n = 2 y_{n-1}$ with $y_1 = x_1 + 3 = 5$. So $y_n = 5 	imes 2^{n-1}$, which means $x_n = 5 	imes 2^{n-1} - 3$. We want $x_n > 10^6$: $$5 	imes 2^{n-1} - 3 > 10^6 \implies 5 	imes 2^{n-1} > 1000003 \implies 2^{n-1} > 200000.6$$ Powers of 2: - $2^{17} = 131072$ - $2^{18} = 262144 > 200000.6$ Thus, $n - 1 = 18 \implies n = 19$.

Answer

No other option is correct for this question.

Recursive Sequence Growth

Explanation

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