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

Given that, on day 1, there are 2 organisms. 
On day-2, there are $2 \times 2 + 3 = 7$
and on day-3, there are $2 \times 7 + 3 = 17$...

Let us try to form a pattern:
$2 = 2 + 0$     $(n = 1)$
$7 = 4 + 3$     $(n = 2)$
$17 = 8 + 9$     $[8 + 3 \times 3]$     $(n = 3)$
$37 = 16 + 21$     $[16 + 3 \times 7]$     $(n = 4)$

So, 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.

Now, check for $n = 19$:
$2^{19} + 3(2^{18} - 1) = 2^{19} + 3 \cdot 2^{18} - 3 = 2 \cdot 2^{19} + 2^{18} - 3 = 2^{20} + 2^{18} - 3$
which is more than 1 million.

Now, check for $n = 18$:
$2^{18} + 3(2^{17} - 1) = 2^{18} + 3 \cdot 2^{17} - 3 = 2 \cdot 2^{18} + 2^{17} - 3 = 2^{19} + 2^{17} - 3$
which is not more than 1 million.

Therefore, $n = 19$