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 $2\times2 + 3 = 7$
and on day-3, there are $2\times7 +3 = 17....$
Let us try to form a pattern:
$2=2+0$      $(n=1)$
$7=4+3$       $(n = 2)$
$17=8+9\ \ \ \  [8+3\times3]$        $(n = 3)$
$37=16+21\ \ \ \ [16 + 3\times7]$    $(n = 4)$
$T(n) = 2^n + 3 (2^{n-1} − 1)$
We know that,  $2^{20}= 2^{10} × 2^{10}$ $=1024 \times 1024$, which is more than $1$ million.
Now, check for $n = 19$,
$2^{19} +3 (2^{18}− 1) = 2^{19}+ 3 ⋅ 2^{18} − 3 = 2 · 2^{19} + 2^{18} − 3 = 2^{20} +2^{18} – 3$, which is more than 1 million.
Now, check for $n = 18$,
$2^{18} + 3 (2^{17} − 1) = 2^{18}+ 3 ⋅ 2^{17} − 3 = 2 · 2^{18} + 2^{17} − 3 = 2^{19} + 2^{17} – 3$, which is not more than a million.
$⇒ n = 19$

So, the answer is $19$.