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

[asy]
unitsize(1 cm);

int i, j;

for (i = 0; i <= 4; ++i) {
draw((i,0)--(i,-4.5));
draw((0,-i)--(4.5,-i));
}

label("\$1\$", (0.5,-0.5));
label("\$a\$", (1.5,-0.5));
label("\$a^2\$", (2.5,-0.5));
label("\$a^3\$", (3.5,-0.5));

label("\$b\$", (0.5,-1.5));
label("\$ab\$", (1.5,-1.5));
label("\$a^2 b\$", (2.5,-1.5));
label("\$a^3 b\$", (3.5,-1.5));

label("\$b^2\$", (0.5,-2.5));
label("\$ab^2\$", (1.5,-2.5));
label("\$a^2 b^2\$", (2.5,-2.5));
label("\$a^3 b^2\$", (3.5,-2.5));

label("\$b^3\$", (0.5,-3.5));
label("\$ab^3\$", (1.5,-3.5));
label("\$a^2 b^3\$", (2.5,-3.5));
label("\$a^3 b^3\$", (3.5,-3.5));

label("\$\dots\$", (5,-2));
label("\$\vdots\$", (2,-5));
[/asy]

(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.

[asy]
unitsize(1 cm);

int i, j;

for (i = 0; i <= 3; ++i) {
for (j = 0; j <= 3; ++j) {
if ((i + j) % 2 == 0) {
fill(shift((i,-j))*((0,0)--(1,0)--(1,-1)--(0,-1)--cycle),black);
}
}}

fill((0,-4)--(1,-4)--(1,-4.5)--(0,-4.5)--cycle,black);
fill((2,-4)--(3,-4)--(3,-4.5)--(2,-4.5)--cycle,black);
fill((4,0)--(4,-1)--(4.5,-1)--(4.5,0)--cycle,black);
fill((4,-2)--(4,-3)--(4.5,-3)--(4.5,-2)--cycle,black);
fill((4,-4)--(4.5,-4)--(4.5,-4.5)--(4,-4.5)--cycle,black);

for (i = 0; i <= 4; ++i) {
draw((i,0)--(i,-4.5));
draw((0,-i)--(4.5,-i));
}

label("\$1\$", (0.5,-0.5), white);
label("\$a\$", (1.5,-0.5));
label("\$a^2\$", (2.5,-0.5), white);
label("\$a^3\$", (3.5,-0.5));

label("\$b\$", (0.5,-1.5));
label("\$ab\$", (1.5,-1.5), white);
label("\$a^2 b\$", (2.5,-1.5));
label("\$a^3 b\$", (3.5,-1.5), white);

label("\$b^2\$", (0.5,-2.5), white);
label("\$ab^2\$", (1.5,-2.5));
label("\$a^2 b^2\$", (2.5,-2.5), white);
label("\$a^3 b^2\$", (3.5,-2.5));

label("\$b^3\$", (0.5,-3.5));
label("\$ab^3\$", (1.5,-3.5), white);
label("\$a^2 b^3\$", (2.5,-3.5));
label("\$a^3 b^3\$", (3.5,-3.5), white);

label("\$\dots\$", (5,-2));
label("\$\vdots\$", (2,-5));
[/asy]
Find the sum of all the numbers on the black squares.

Dec 10, 2019

#1
0

GET  REAL !

Dec 10, 2019
#2
0

I'm sorry, but I don't understand the question.

Dec 10, 2019
#3
0

I don't know how to upload PNGs so this is the only way I can show you the grid.

The image is really dark but you can turn up the brightness to see

part a grid ------>     https://latex.artofproblemsolving.com/8/c/1/8c1cac67360acec4ef623054fed4ef33fd6c381a.png

part b grid ------>       https://latex.artofproblemsolving.com/e/3/9/e3959697cb9ff1295895297696a62834311cd76f.png

Dec 10, 2019
#4
+1

(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