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The real numbers a and b satisfy |a| < 1 and |b| < 1. (a) In a grid that extends infinitely, the first row contains the numbers 1, a, a^2, . . . . The second row contains the numbers b, ab, a^2 b, . . . . In general, each number is multiplied by a to give the number to the right of it, and each number is multiplied by b to give the number below it. Find the sum of all numbers in the 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 that lie on the black squares.

 May 3, 2020
 #1
avatar+20914 
+1

First problem:

 

The first row is a geometric series with first term  1  and common ratio  a  whose sum is  1/(1 - a).

 

The second row is also geometric with first term  b  and common ratio  a  whose sum is  b/(1 - a).

 

The third row is also geometric with first term  b2  and common ratio  a  whose sum is  b2/(1 - a).

 

The sum of the rows is:  1/(1 - a)  +  b/(1 - a)  +  b2/(1 - a)  + ...  =   1/(1 - a) · [ 1 + b + b2 + ... ]

 

The series  1 + b + b2 + ...  is geometric with first term  1  and common ratio  b  whose sum is  1/(1 - b).

 

Therefore, the sum of the rows is:  1/(1 - a) · [ 1 + b + b2 + ... ]  =  1/(1 - a) · 1/(1 - b)  =  1 / [ (1 - a)(1 - b) ]

 May 3, 2020
 #2
avatar+20914 
+1

Second problem:

 

1 + a2 + a4 + ...  =  1/(1 - a2)

ab + a3b + a5b + ...  =  ab/(1 - a2)

b2 + a2b2 + a4b2 + ...  =  b2/(1 - a2)

ab3 + a3b3 + a5b3 + ...  =  ab3/(1 - a2)

...

Sum  =   1/(1 - a2)  +  ab/(1 - a2)  +  b2/(1 - a2)  +  ab3/(1 - a2)  +  ...

         =   1/(1 - a2) · [ 1 + ab + b2 + ab3 + b4 + ab5 + b6 + ab7 + ...]

         =   1/(1 - a2)   [ (1 + b2 + b4 + b6 + ...)  +  (ab + ab3 + ab5 + ab7 + ...) ]

         =   1/(1 - a2) · [ ( 1/(1 - b2 ) + ( ab/(1 - b2) ) ]

         =   [ 1/(1 - a2) · 1/(1 - b2 ) ] · [ 1 + ab ]

         =  ( 1 + ab ) / [ (1- a2)(1 - b2) ] 

 May 3, 2020

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