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Let \(a\) and \(b\) be real numbers, where \(|a| < 1\) and \(|b| < 1.\) In an infinite grid, I write the numbers \(1, \ a, \ a^2, \ a^3, \ \dots\)
in the first row. After that, each number is equal to \(b\) times the number above it. For example, the numbers in the second row are \(b, \ ab, \ a^2 b, \ a^3 b, \ \dots.\)

(a) Find the sum of all the numbers in the infinite grid.

(b) Now suppose the grid is colored like a chessboard, with alternating black and white squares, as shown below.

Find the sum of all the numbers on the black squares.

 Dec 9, 2019
 #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, 2019
 #2
avatar+33 
+1

Did you mean to add a slash here? 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))

                                                                                                                                     ^

Divineology  Dec 10, 2019
 #3
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+1

Yes, there should be a slash there.

Guest Dec 10, 2019
 #4
avatar+33 
+1

Ok, I just wanted to make sure! cool

Divineology  Dec 13, 2019

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