Question 471020
Here's the basic table you should have:

Year x: Beginning --> End
--------------------------
year 1: 1000 --> 2000
year 2: 2000 --> 4000
year 3: 4000 --> 8000
year 4: 8000 --> 16000
year 5: 16000 --> 32000
year 6: 32000 --> 64000
year 7: 64000 --> 128000
year 8: 128000 --> 256000
year 9: 256000 --> 512000
year 10: 512000 --> 1024000

Sum = 1000+2000+4000+8000+16000+32000+64000+128000+256000+512000+1024000 = 2047000 



So you are correct. Good job.



A shortcut is to realize that these values form a geometric sequence that fit the form {{{a[n]=a*r^(n-1)}}} where n starts at n=1. In your case, {{{a=1000}}} and {{{r=2}}}. So the formula is {{{a[n]=1000*2^(n-1)}}}



Now you can use the formula {{{S=a((1-r^n)/(1-r))}}}, where S is the partial sum from n=1 up to the nth term, which in this case is n=11 (note: at the END of year 10 is the start of year 11, so we sum up to 11 to include the last readers taught in year 10).



So this means that the partial sum is {{{S=1000((1-2^11)/(1-2))=1000((1-2048)/(-1))=1000((-2047)/(-1))=1000(2047)=2047000}}}



So we get the same answer.