A woman labels the squares of a very large chessboard 1 through 64. On each square k, the woman puts 2^k grains of rice. How many more grains of rice are placed on the 10th square than on the first 8 squares combined?
We can solve this with patterns!
Notice how 2^2=2^1+2^0+1
2^3=2^2+2^1+2^0+1
Etc...
So this means The first k squares al added together is equal the the k+1 square - 1.
With this, the first 8 squares added together is 2^9-1 = 511, 2^10=1024, 1024-511=513
Good work Babada,
I did it using GP formula and I got just 1 different from you.
this is ok because it us up to our guest asker to look at our working and decide who is correct :)
\(2^1+2^2+ ....+2^8 = \frac{a(r^n-1)}{r-1}=\frac{2(2^8-1)}{2-1}=2^9-2\)
\(2^{10}-(2^9-2)\\ =2^{10}-2^9+2\\ =2^9*(2-1)+2\\ =2^9+2\\ =512+2\\ =514 \)