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 nth day exceeds one million, then the lowest possible value of n is

Answer 1

On day 1, there are 2 organisms. On day 2, there are $2 \times 2 + 3 = 7$ organisms. On day 3, there are $2 \times 7 + 3 = 17$ organisms.

Let's identify a pattern:

• $n = 1$: $2 = 2 + 0$
• $n = 2$: $7 = 4 + 3$
• $n = 3$: $17 = 8 + 9$ (which is $8 + 3 \times 3$)
• $n = 4$: $37 = 16 + 21$ (which is $16 + 3 \times 7$)

The general term is: $T(n) = 2^n + 3(2^{n-1} - 1)$

We know that $2^{20} = 2^{10} \times 2^{10} = 1024 \times 1024$, which is more than 1 million.

Checking for $n = 19$:

$2^{19} + 3(2^{18} - 1) = 2^{19} + 3 \cdot 2^{18} - 3 = 2 \cdot 2^{19} + 2^{18} - 3 = 2^{20} + 2^{18} - 3$. This value is more than 1 million.

Checking for $n = 18$:

$2^{18} + 3(2^{17} - 1) = 2^{18} + 3 \cdot 2^{17} - 3 = 2 \cdot 2^{18} + 2^{17} - 3 = 2^{19} + 2^{17} - 3$. This value is not more than 1 million.

Therefore, $n = 19$