Question 29590
There are 64 squares
the farmer put 1 grain in a corner square
2 in the next,
4 in the next,
8 in the next
let g = total number of grains
let n = number of squares
the rule is
g = 2^0 + 2^1 + 2^2 + 2^3 + . . . + 2^n
so after, let's say 6 squares, the number of grains on the checkerboard is
{{{g = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5}}}
(anything to the 0 power is one)
{{{g = 1 + 2 + 4 + 8 + 16 + 32}}}
the total is 63
the totals are always 2^(n+1) - 1
so if n = 64
{{{g = 2^0 + 2^1 + 2^2 + 2^3 + . . . + 2^64}}}
and the sum g is
{{{g = 2^65 - 1}}}
to the limits of my calculator, I get
3689348815 . . . and 10 more decimal places or
{{{3.689348815 * 10^19}}}