Exceeding 2 000 000!

Sum to $n$ terms is

$\color{#3D99F6}Substitute\ a =1, r=2\ into\ S_n=\frac{a(r^n -1)}{r - 1}$ $\large S_n = 1\frac{(2^n - 1)}{2 - 1} =\ 2^n - 1$

If this is to exceed 2 000 000 then:

$S_n > 2 000 000$

$2^n - 1 > 2 000 000$

$2^n > 2 000 001$ $\color{#3D99F6}Add\ 1$

$n log(2) > log(2 000 001)$

$n > \frac{log(2 000 001)}{log(2)}$ $\color{#3D99F6}Use\ law\ of\ logs$

$\large \color{#20A900}n > 20.9$

It needs 21 terms to exceed 2 000 000
$\color{#3D99F6}Round\ the\ up\ n\ to\ the\ nearest\ integer$ $\large {\color{#D61F06}n = 21}$

The nth term is $2^n$ and the sum of n terms is $2^{n+1}-1$ . Now solve the equation $2^{n+1}-1=2000000$ . This gives $n = log_2(2000001)\approx 21$ .