Question

Can Anybody solve this - in a chess board if I put one coin for 1 square of chessboard , two coins for next square , four coins for the next , 8 for the next and so on for all 64 squares with each having double the number of coins as the square before . find the total number of coins I will get .

Plz help me to solve this sum

Answer 1

Answer:

Total number of coins he will get $=2^{64}-1$

Step-by-step explanation:

No. of coins in square-1 $=1$

No. of coins in square-2 $=1(2)= 2$

No. of coins in square-3 $=2(2) =2^2$

No. of coins in square-4 $=2(2^2) =2^3$

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No. of coins in square-64 $=2(2^{62}) =2^{63}$

clealy the numbers $1, 2, 2^2, 2^3........2^{63}$ form a G.P

Total number of coins he will get

$=\frac{a(r^n-1)}{r-1}$

$=\frac{1(2^{64}-1)}{2-1}$

$=\frac{1(2^{64}-1)}{1}$

$=2^{64}-1$