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 $1,$ $a,$ $a^2,$ $\dots.$ The second row contains the numbers $b,$ $ab,$ $a^2 b,$ $\dots.$ 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.
[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]
(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.
[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]
Sorry but I don't know how to show the grids.
each row is a geometric progression. Starting with n=0, row n is the GP
b^n (1 + a + a^2 + ...)
so the sum Sn of row n is
b^n * 1/(1-a)
Now you have another GP, which is the sum of all the row sums. That is
1/(1-a) (1 + b + b^2 + ...) whose sum is 1/(1-a) * 1/(1-b)
So the grand total of the whole grid is 1/((1-a)(1-b))
a)
We start by adding the first row of the grid. We set $S$ as $1+a+a^2+a^3...$ Then we set $a(S)$ as $a+a^2+a^3+a^4...$ because $a$ is the common ratio. From this we can get $(a-1)S = -1$ which becomes $S = \frac{-1}{a-1}.$ Changing the top part to $1$ we get $\frac{1}{1-a}.$ We observe that the second row of the grid is the Same as the first row of the grid except that the second row's terms are $b$ times more than the first row. We can easily tell that the second row adds up to $\frac{1}{1-a}*b$ which is $\frac{b}{1-a}.$ The entire grid's sum is therefore $\frac{1}{1-a}+\frac{b}{1-a}+\frac{b^2}{1-a}...$ which simplifies to $\frac{1}{1-a}(\frac{1}{1-a}+\frac{b}{1-a}+\frac{b^2}{1-a}...)$ Our final answer is $\frac{1(b^n)}{b-1}*\frac{1}{1-a}$ which is $\frac{1}{1-b}*\frac{1}{1-a},$ therefore giving us $1-a-b+ba.$
a)
We start by adding the first row of the grid. We set $S$ as $1+a+a^2+a^3...$
Then we set $a(S)$ as $a+a^2+a^3+a^4...$ because $a$ is the common ratio.
From this, we can get $(a-1)S = -1$ which becomes $S = \frac{1}{1-a}.$
We observe that the second row of the grid is the same as the first row of the grid except that the second row's terms are $b$ times more than the first row.
We can easily tell that the second row adds up to $\frac{1}{1-a}*b$ which is $\frac{b}{1-a}.$
The entire grid's sum is therefore $\frac{1}{1-a}(1+b+b^2...)$
Similar to $1+a+a^2+a^3... = \frac{1}{1-a},$ we can also state the sentence $1+b+b^2+b^3... = \frac{1}{1-b}.$
Our final answer is $\frac{1}{1-b}*\frac{1}{1-a}=\frac{1}{1-a-b+ba}.$
b)
We start by the same approach as problem a) by adding each black square in each row from the grid.
The first row is $1+a^2+a^4... = \frac{1}{1-a^2}$ which is the sum of the first row.
The second row is $ab+a^3b+a^5b... = ab*(1+a^2+a^4...) = \frac{1}{1-a^2}*ab = \frac{ab}{1-a^2}.$
The third row is $b^2+a^2b^2+a^4b^2... = b^2*(1+a^2+a^4...) = \frac{1}{1-a^2}*b^2 = \frac{b^2}{1-a^2}.$
The fourth row is $ab^3+a^3b^3+a^5b^3... = ab^3*(1+a^2+a^4...) = \frac{1}{1-a^2}*ab^3 = \frac{ab^3}{1-a^2}.$
The fifth row is $b^4+a^2b^4+ a^4b^4... = b^4*(1+a^2+a^4...) = \frac{1}{1-a^2}*b^4 = \frac{b^4}{1-a^2}$
The sixth row is $ab^5+a^3b^5+a^5b^5... = ab^5(1+a^2+a^4...) = \frac{1}{1-a^2}*ab^5 = \frac{ab^5}{1-a^2}$
We can organize these terms to two sums. The first sum is the sum of the odd rows. The second sum is the sum of the even rows.
The first sum is $\frac{1}{1-a^2}+\frac{b^2}{1-a^2}+\frac{b^4}{1-a^2}... = \frac{1}{1-a^2}(1+b^2+b^4...) = \frac{1}{1-a^2}(\frac{1}{1-b^2}).$
The second sum is $\frac{ab}{1-a^2}+\frac{ab^3}{1-a^2}+\frac{ab^5}{1-a^2}... = \frac{ab}{1-a^2}(1+b^2+b^4...) = \frac{ab}{1-a^2}(\frac{1}{1-b^2})$
The total sum is therefore, $\frac{1 + ab}{(1- a^2)(1 - b^2)} $
No problem! I'll help you solve the problem.
(a) To find the sum of all numbers in the grid, we can start by finding the sum of the numbers in each column. In the first column, the sum is $1 + b + b^2 + b^3 + \dots = \frac{1}{1 - b}$. Similarly, the sum of the numbers in the second column is $a + ab + a^2b + a^3b + \dots = \frac{a}{1 - a}$. Continuing this pattern, the sum of the numbers in the $n$th column is $\frac{a^n}{1 - a}$.
Now, to find the sum of all numbers in the grid, we can sum the geometric series for each column. Since $|a| < 1$, the series $\frac{a^n}{1 - a}$ converges, and we can use the formula for the sum of an infinite geometric series. Thus, the sum of all numbers in the grid is
\[\frac{1}{1 - b} + \frac{a}{1 - a} + \frac{a^2}{1 - a} + \dots = \frac{1}{1 - b} + \frac{a}{1 - a} \cdot \frac{1}{1 - a} = \frac{1}{(1 - b)(1 - a)}.\]
(b) To find the sum of all the numbers that lie on the black squares, we can consider that each black square is the intersection point of two white squares. This means that each black square is the product of the number to its right and the number below it. Since $a$ and $b$ are both less than $1$ in absolute value, the product of any two numbers in the grid will be smaller in absolute value than either of the two numbers being multiplied. Therefore, the sum of all the numbers on the black squares will be smaller in absolute value than the sum of all the numbers on the white squares.
So, the sum of all the numbers on the black squares is also $\frac{1}{(1 - b)(1 - a)}$.
I hope that helps! Let me know if you have any other questions.
No problem! I can help you solve the problem without needing to see the grids.
(a) To find the sum of all numbers in the grid, we can notice that the numbers in each column form a geometric sequence. The first term of the sequence is the number in the first row of that column, and the common ratio is $a$. Similarly, the numbers in each row form a geometric sequence, with the first term being the number in the first column of that row, and the common ratio being $b$.
Using the formula for the sum of a geometric series, the sum of the numbers in each column is given by:
\[1 + a + a^2 + \dots = \frac{1}{1-a}\]
And the sum of the numbers in each row is given by:
\[1 + b + b^2 + \dots = \frac{1}{1-b}\]
Since the grid extends infinitely in both directions, the sum of all numbers in the grid can be found by adding up the sums of each column. To do this, we can sum the geometric series for each column by multiplying the sum of the numbers in each column by the common ratio $a$, and then add them up:
\[(1 + a + a^2 + \dots) + (a)(1 + a + a^2 + \dots) + (a^2)(1 + a + a^2 + \dots) + \dots\]
This simplifies to:
\[(1 + a + a^2 + \dots)(1 + a + a^2 + \dots)\]
Using the formula for the product of two geometric series, this can be further simplified to:
\[\frac{1}{(1-a)(1-a)} = \frac{1}{(1-a)^2}\]
Therefore, the sum of all numbers in the grid is $\frac{1}{(1-a)^2}$.
(b) To find the sum of all the numbers that lie on the black squares, we can use the same logic as in part (a), but consider only the black squares.
Notice that the black squares form a smaller grid within the larger grid. This smaller grid includes every other number in each row and each column of the original grid. We can represent this smaller grid as a $2\times 2$ grid:
\[
\begin{bmatrix}
1 & a \\
b & ab \\
\end{bmatrix}
\]
Using the same approach as in part (a), the sum of the numbers in this smaller grid is given by:
\[(1 + a)(1 + b)\]
So, the sum of all the numbers that lie on the black squares in the original grid is $(1 + a)(1 + b)$.