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Could someone please help we with this problem and give me an explanation for how they get to their solution? Thank you!

 Dec 10, 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)).

 Dec 10, 2020

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