my friend and myself played chess and i won..he had promised me to give watever i want if i win and fortunately i managed to win..i asked for 1 chocolate for the first square of the chess board, 2 for the second,4 for the third and so on..how many chocolates does he owe me??
2007-08-03 06:13:00 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
2^64 - 1
there are 64 spots.
First one: 1
Second one: 1 + 2 = 3 (2^2 - 1)
Third one: 1 + 2 + 4 = 7 (2^3 - 1)
...
At the 64th, your sum will be 2^64 - 1.
2007-08-03 06:17:08 · answer #1 · answered by Anonymous · 0⤊ 0⤋
This is a gemetric series with r = 2
The sum of the 1st n terms is (1-r^n)/(1-r)
So sum = (1-2^64)/(1-2) = 2^64-1
2007-08-03 13:26:18 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
1 + 2 + 4 + .... + 2^63
= 2^64 - 1.
2007-08-03 13:19:09 · answer #3 · answered by Anonymous · 0⤊ 0⤋
2⁶⁵-1 roughly... very roughly... 36.8 quintillion chocolates. You might want to go into business.... except it would take a while for all the people in the world to eat that much chocolate.
2007-08-03 13:28:18 · answer #4 · answered by gugliamo00 7 · 0⤊ 0⤋
1.8E19 chocolates
2007-08-03 13:23:16 · answer #5 · answered by Alberd 4 · 0⤊ 0⤋