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+1
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I am completely stuck on this. I can tell that each row is a geometric series. But what should I do?

 Jun 30, 2020
 #1
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(a) The sum of the numbers in the first row is 1/(1 - a).

 

The sum of the numbers in the second row is ab/(1 - a).

 

The sum of the numbers in the third row is (a^2 b^2)/(1 - a).

 

So, the sum of the numbers in the rows form a geometric sequence, which adds up to

1/(1 - a) + ab/(1 - a) + (a^2 b^2)/(1 - a) + ...  = 1/((1 - a)(1 - ab)).

 

(b) Since the colors of the chessboard alternate white and black, the sum of the numbers on the black squares is equal to the sum of the numbers on the white squares, except for the numbers that are on every other white square, and every other black square.

 

The sum of the numbers that are on every other white square is a/((1 - a)(1 - ab)), and the sum of the numbers that are on every other black square is b/((1 - a)(1 - ab)), so to find the sum of the numbers on the black squares, we take half the difference, which gives us a sum of

 

1/((1 - a)(1 - ab)) + 1/2*a/((1 - a)(1 - ab)) + 1/2*b((1 - a)(1 - ab)) = (a + b + 2)/(2(1 - a)(1 - ab)).

 Jun 30, 2020

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