+306 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 \(b, ab, a^2b, ....\) 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 the 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.
I couldn't upload any images in this problem, so here are links to the images:
a) https://www.flickr.com/photos/189314996@N03/50113679286/in/dateposted-public/
b) https://www.flickr.com/photos/189314996@N03/50113680186/in/dateposted-public/
(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)).
+306 Could you please explain how you found those sums? I don't quite understand.
After working on this problem for a while, I found that the sum of the first row is \(\frac{1}{1-a}\) because the formula for the sum of a geometric series is \(\frac{\text{first term}}{1-r}\), where r=the common ratio. Similarly, I found that the sum of the second row is \(\frac{b}{1-a}\), and the sum of the third row is \(\frac{b^2}{1-a}\). This led me to believe that the sum of each row could be multiplied by b to get the sum of the next row, and therefore, the sums of the rows form a geometric series. I then found the sum of that series in the same way. Since the sum of a geometric series is \(\frac{\text{first term}}{1-r}\), I found that the sum of all the squares in the grid is \(\frac{\frac{1}{1-a}}{1-b}=\frac{1}{(1-a)(1-b)}\). I haven't worked on part b yet, but I assume the sum can be found the same way.
The difference in our solutions has me wondering if I made a mistake, and if so, where I made it. I would really appreciate if someone could help me out with this!
