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Let a and 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 b, ab, a^2b, a^3b.

 

Here's the rest of the grid:


https://latex.artofproblemsolving.com/8/c/1/8c1cac67360acec4ef623054fed4ef33fd6c381a.png

 

(a) Find the sum of all the numbers on the infinite grid.

 

Part 2

 

https://latex.artofproblemsolving.com/e/3/9/e3959697cb9ff1295895297696a62834311cd76f.png

 

(b) Find the sum of all the numbers on the black squares. (|a| < 1 and |b| < 1)

 

Thank you so much in advance!

 Dec 10, 2019
 #1
avatar+23905 
+2

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 b times the number above it. For example, the numbers in the second row are b, ab, a^2b, a^3b.

 

Here's the rest of the grid:


 

(a) Find the sum of all the numbers on the infinite grid.
Infinite sum:

\(\sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{\infty} a^k b^l \right) \)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{\infty} a^k b^l \right)} \\ &=& \sum \limits_{l=0}^{\infty} b^l \left( \sum \limits_{k=0}^{\infty} a^k \right) \quad | \quad \sum \limits_{k=0}^{\infty} a^k = \dfrac{1}{1-a} \\ &=& \sum \limits_{l=0}^{\infty} b^l \left( \dfrac{1}{1-a} \right) \\ &=& \left( \dfrac{1}{1-a} \right) \sum \limits_{l=0}^{\infty} b^l \quad | \quad \sum \limits_{l=0}^{\infty} b^l = \dfrac{1}{1-b} \\ &=& \left( \dfrac{1}{1-a} \right) \left( \dfrac{1}{1-b} \right) \\ &=& \mathbf{\dfrac{1}{(1-a)(1-b)}} \\ \hline \end{array}\)

 

laugh

 Dec 10, 2019
 #2
avatar+23905 
+2

Part 2

 

(b) Find the sum of all the numbers on the black squares. (|a| < 1 and |b| < 1)

 

Sum of all the numbers on the black squares:

\(1 \\ +a^2+ab+b^2 \\ +a^4+a^3b+a^2b^2+ab^3+b^4 \\ +a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6 \\ +\ldots\)

 

Infinite sum:
\(\sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{2l} a^{2l-k}b^k \right) \)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{2l} a^{2l-k}b^k \right)} \\ &=& \sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{2l} a^{2l}a^{-k}b^k \right) \\ &=& \sum \limits_{l=0}^{\infty} a^{2l}\left( \sum \limits_{k=0}^{2l} a^{-k}b^k \right) \\ &=& \sum \limits_{l=0}^{\infty} a^{2l} \left( \sum \limits_{k=0}^{2l} \left(\dfrac{b}{a}\right)^k \right) \quad | \quad \sum \limits_{k=0}^{2l} \left(\dfrac{b}{a}\right)^k = \dfrac{1- \left(\dfrac{b}{a}\right)^{2l+1} }{1-\dfrac{b}{a} } \\ &=& \sum \limits_{l=0}^{\infty} a^{2l} \left( \dfrac{1- \left(\dfrac{b}{a}\right)^{2l+1} }{1-\dfrac{b}{a} } \right) \\ &=& \sum \limits_{l=0}^{\infty} a^{2l} \left( \dfrac{a- \left(\dfrac{b}{a}\right)^{2l}b }{a-b } \right) \\ &=& \dfrac{1}{(a-b)} \sum \limits_{l=0}^{\infty} a^{2l} \left( a- \left(\dfrac{b}{a}\right)^{2l}b \right) \\ &=& \dfrac{1}{(a-b)} \sum \limits_{l=0}^{\infty} \left( a^{2l+1}- \left(\dfrac{ a^{2l}b^{2l}}{a^{2l}}\right)b \right) \\ &=& \dfrac{1}{(a-b)} \sum \limits_{l=0}^{\infty} \left( a^{2l+1}- b^{2l+1} \right) \\ &=& \dfrac{1}{(a-b)} \left( \sum \limits_{l=0}^{\infty} \left( a^{2l+1} \right) - \sum \limits_{l=0}^{\infty} \left( b^{2l+1} \right) \right) \\ &=& \dfrac{1}{(a-b)} \left( a\sum \limits_{l=0}^{\infty} a^{2l} - b\sum \limits_{l=0}^{\infty} b^{2l} \right) \quad | \quad \sum \limits_{l=0}^{\infty} a^{2l} = \dfrac{1}{1-a^2},\ \sum \limits_{l=0}^{\infty} b^{2l} = \dfrac{1}{1-b^2} \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{a}{1-a^2} - \dfrac{b}{1-b^2} \right) \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{a(1-b^2)-b(1-a^2)}{(1-a^2)(1-b^2)} \right) \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{a-ab^2-b+ba^2}{(1-a^2)(1-b^2)} \right) \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{(a-b)+ab(a-b)}{(1-a^2)(1-b^2)} \right) \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{(a-b)(1+ab)}{(1-a^2)(1-b^2)} \right) \\ &=& \mathbf{ \dfrac{1+ab}{(1-a^2)(1-b^2)} } \\ \hline \end{array}\)

 

laugh

 Dec 10, 2019

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