Question

Can someone help me with this

Answer 1

Answer:

19 m

Step-by-step explanation:

This is Geometric progression with

$\left \{ {{u_1=1,\: u_2 =2} \atop {u_n=2^{n-1}u_1} \right.$

Then the fomular of sum is

$S=u_1\frac{2^n-1}{2-1} =1\times(2^n-1)=2^n-1\\To\: save\: 1 \:million\: then \; our \:sum \: will \: be\: 1000000\\Hence,\\2^n-1=1000000\\2^n=1000001\\n=log_{2}{1000001}\approx19,9\\Thus, it\:would \;be \:19 \:months \:before\: we\: saved \:1000000 \:pounds$