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))
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.
3 answers
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 $1-a-b+ba.$
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)} $
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)} $
1 answer