- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

$b_1=3$, $b_2=3$, and for $n \ge 2$, $b_{n+1}b_{n-1}=b_n^2+2007$.

Find the largest integer less than or equal to $\dfrac{b_{2007}^2+b_{2006}^2}{b_{2007}b_{2006}}$.

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

$b_1=3$, $b_2=3$, and for $n \ge 2$, $b_{n+1}b_{n-1}=b_n^2+2007$.

Find the largest integer less than or equal to $\dfrac{b_{2007}^2+b_{2006}^2}{b_{2007}b_{2006}}$.

- Thread starter
- Admin
- #2

- Feb 14, 2012

- 3,909

Build a table for a few terms for $n>2$ to check if there is a pattern to be observed:

n | $b_{n+1}=\dfrac{b_n^2+2007}{b_{n-1}}$ | $\dfrac{b_{n+1}}{b_n}$ |

2 | $b_3=\dfrac{b_2^2+2007}{b_1}=\dfrac{3^2+2007}{3}=672$ | - |

3 | $b_4=\dfrac{b_3^2+2007}{b_2}=\dfrac{672^2+2007}{3}=151197$ | $\dfrac{b_4}{b_3}=\dfrac{151197}{672} \approx 224.9955357 \approx 225$ |

4 | $b_5=\dfrac{b_4^2+2007}{b_3}=\dfrac{151197^2+2007}{672}=34018653$ | $\dfrac{b_5}{b_4} \approx 224.99555 \approx 225$ |

5 | $b_6=\dfrac{b_5^2+2007}{b_4}=\dfrac{34018653^2+2007}{151197}=7654045728$ | $\dfrac{b_6}{b_5} \approx 224.99556 \approx 225$ |

6 | $b_7=\dfrac{b_6^2+2007}{b_5}=\dfrac{7654045728^2+2007}{34018653}=1.72212627\times10^{12}$ | $\dfrac{b_7}{b_6} \approx 225$ |

7 | $b_8=\dfrac{b_7^2+2007}{b_6}=\dfrac{(1.72212627 \times10^{12})^2+2007}{7654045728}=3.8747 \times10^{14}$ | $\dfrac{b_8}{b_7} \approx 225$ |

$\vdots$ | $\vdots$ | $\vdots$ |

That is, we have generated a "geometric-like" sequence with the common ratio that takes the approximate value of 225 (in fact, the so-called common ratio is increasing as $n$ increases).

Another thing that is worth mentioning here is as $n$ increases, the value of $b_n$ increases monumentally that the addition of the value 2007 in the numerator brings no significant increment to the numerator and we can thus treat the whole equation of $b_{n+1}b_{n-1}=b_n^2+2007$ as $b_{n+1}b_{n-1}=b_n^2$ or $\dfrac{b_{n+1}}{b_n}=\dfrac{b_n}{b_{n-1}}$, i.e. the terms of $b_n$ generate a geometric sequence.

$\therefore \dfrac{b_{2007}^2+b_{2006}^2}{b_{2007}b_{2006}}= \dfrac{b_{2007}}{b_{2006}}+\dfrac{b_{2006}}{b_{2007}}=\dfrac{b_{2007}}{b_{2006}}+\dfrac{1}{\dfrac{b_{2007}}{b_{2006}}}<225+\dfrac{1}{225} <225.00444$

Hence the largest integer less than or equal to $\dfrac{b_{2007}^2+b_{2006}^2}{b_{2007}b_{2006}}$ is 225.

- Moderator
- #3

- Feb 7, 2012

- 2,785

This is a wicked little problem, and I am not sure that we have got to the root of it.

*less* than 225 to ensure that $x_n + x_n^{-1} = \dfrac{b_{n+1}^2+b_{n}^2}{b_{n+1}b_{n}}$ is also less than 225. In that case, "the largest integer less than or equal to $\dfrac{b_{2007}^2+b_{2006}^2}{b_{2007}b_{2006}}$" would be 224, not 225. As far as I can see, that appears to be what actually happens.

The condition $x_n + x_n^{-1} < 225$ is equivalent to the condition that $x_n$ should lie between the two roots of the equation $x + x^{-1} = 225$, or $x^2 - 225x + 1 = 0$. The larger of the two roots is $\frac12\bigl(225 + \sqrt{50621}\bigl) \approx 224.99556.$ So we have to ask whether $x_n$ stays below that value.

Divide both sides of the equation $b_{n+1}b_{n-1}=b_n^2+2007$ by $b_nb_{n-1}$ to get $x_{n+1} = \dfrac{2007}{b_{n-1}b_n} + x_n$. Apply that relation inductively to see that $$x_n = 2007\left(\frac1{b_{n-1}b_n} + \frac1{b_{n-2}b_{n-1}} + \ldots + \frac1{b_4b_3}\right) + x_3.$$ Next, the sequence $(x_n)$ is increasing, and $x_3 = b_4/b_3 \approx 224.9955357 >224.9$. Therefore $x_n = b_{n+1}/b_n >224.9$ whenever $n\geqslant3.$ Thus $b_n > 224.9b_{n-1}$ and (again by an inductive argument) $b_n > 224.9^{n-3}b_3 = 672\cdot 224.9^{n-3}.$ It follows that $$x_n < \frac{2007}{672^2}\left(\frac1{224.9^{2n-7}} + \frac1{224.9^{2n-9}} + \ldots + \frac1{224.9}\right) + x_3.$$ If we replace the geometric series in the large brackets by the corresponding infinite geometric series, which has sum $\dfrac1{224.9(1-224.9^{-1})}$ then we get the estimate $$x_n < \frac{2007}{672^2\cdot 224.9(1-224.9^{-1})} + x_3 \approx 0.0000198 + 224.9955357 = 224.9955555 .$$ This is less than $224.99556$, so it appears that $x_n+x_n^{-1}$ always stays below 225.

I hate to rely on such delicate numerical evidence, especially since my little calculator only works to 8 significant figures. I would much prefer to have an analytic argument showing that the answer to the problem is $224$, but I don't see how to do that.

You have shown that the ratio $x_n = \dfrac{b_{n+1}}{b_n}$ is very close to 225. But suppose it always stays just sufficiently

Given $b_1=3$, $b_2=3$, and for $n \ge 2$, $b_{n+1}b_{n-1}=b_n^2+2007$.

Build a table for a few terms for $n>2$ to check if there is a pattern to be observed:

n $b_{n+1}=\dfrac{b_n^2+2007}{b_{n-1}}$ $\dfrac{b_{n+1}}{b_n}$ 2 $b_3=\dfrac{b_2^2+2007}{b_1}=\dfrac{3^2+2007}{3}=672$ - 3 $b_4=\dfrac{b_3^2+2007}{b_2}=\dfrac{672^2+2007}{3}=151197$ $\dfrac{b_4}{b_3}=\dfrac{151197}{672} \approx 224.9955357 \approx 225$ 4 $b_5=\dfrac{b_4^2+2007}{b_3}=\dfrac{151197^2+2007}{672}=34018653$ $\dfrac{b_5}{b_4} \approx 224.99555 \approx 225$ 5 $b_6=\dfrac{b_5^2+2007}{b_4}=\dfrac{34018653^2+2007}{151197}=7654045728$ $\dfrac{b_6}{b_5} \approx 224.99556 \approx 225$ 6 $b_7=\dfrac{b_6^2+2007}{b_5}=\dfrac{7654045728^2+2007}{34018653}=1.72212627\times10^{12}$ $\dfrac{b_7}{b_6} \approx 225$ 7 $b_8=\dfrac{b_7^2+2007}{b_6}=\dfrac{(1.72212627 \times10^{12})^2+2007}{7654045728}=3.8747 \times10^{14}$ $\dfrac{b_8}{b_7} \approx 225$ $\vdots$ $\vdots$ $\vdots$

That is, we have generated a "geometric-like" sequence with the common ratio that takes the approximate value of 225 (in fact, the so-called common ratio is increasing as $n$ increases).

Another thing that is worth mentioning here is as $n$ increases, the value of $b_n$ increases monumentally that the addition of the value 2007 in the numerator brings no significant increment to the numerator and we can thus treat the whole equation of $b_{n+1}b_{n-1}=b_n^2+2007$ as $b_{n+1}b_{n-1}=b_n^2$ or $\dfrac{b_{n+1}}{b_n}=\dfrac{b_n}{b_{n-1}}$, i.e. the terms of $b_n$ generate a geometric sequence.

$\therefore \dfrac{b_{2007}^2+b_{2006}^2}{b_{2007}b_{2006}}= \dfrac{b_{2007}}{b_{2006}}+\dfrac{b_{2006}}{b_{2007}}=\dfrac{b_{2007}}{b_{2006}}+\dfrac{1}{\dfrac{b_{2007}}{b_{2006}}}<225+\dfrac{1}{225} <225.00444$

Hence the largest integer less than or equal to $\dfrac{b_{2007}^2+b_{2006}^2}{b_{2007}b_{2006}}$ is 225.

The condition $x_n + x_n^{-1} < 225$ is equivalent to the condition that $x_n$ should lie between the two roots of the equation $x + x^{-1} = 225$, or $x^2 - 225x + 1 = 0$. The larger of the two roots is $\frac12\bigl(225 + \sqrt{50621}\bigl) \approx 224.99556.$ So we have to ask whether $x_n$ stays below that value.

Divide both sides of the equation $b_{n+1}b_{n-1}=b_n^2+2007$ by $b_nb_{n-1}$ to get $x_{n+1} = \dfrac{2007}{b_{n-1}b_n} + x_n$. Apply that relation inductively to see that $$x_n = 2007\left(\frac1{b_{n-1}b_n} + \frac1{b_{n-2}b_{n-1}} + \ldots + \frac1{b_4b_3}\right) + x_3.$$ Next, the sequence $(x_n)$ is increasing, and $x_3 = b_4/b_3 \approx 224.9955357 >224.9$. Therefore $x_n = b_{n+1}/b_n >224.9$ whenever $n\geqslant3.$ Thus $b_n > 224.9b_{n-1}$ and (again by an inductive argument) $b_n > 224.9^{n-3}b_3 = 672\cdot 224.9^{n-3}.$ It follows that $$x_n < \frac{2007}{672^2}\left(\frac1{224.9^{2n-7}} + \frac1{224.9^{2n-9}} + \ldots + \frac1{224.9}\right) + x_3.$$ If we replace the geometric series in the large brackets by the corresponding infinite geometric series, which has sum $\dfrac1{224.9(1-224.9^{-1})}$ then we get the estimate $$x_n < \frac{2007}{672^2\cdot 224.9(1-224.9^{-1})} + x_3 \approx 0.0000198 + 224.9955357 = 224.9955555 .$$ This is less than $224.99556$, so it appears that $x_n+x_n^{-1}$ always stays below 225.

I hate to rely on such delicate numerical evidence, especially since my little calculator only works to 8 significant figures. I would much prefer to have an analytic argument showing that the answer to the problem is $224$, but I don't see how to do that.

Last edited:

- Thread starter
- Admin
- #4

- Feb 14, 2012

- 3,909

Yes, you're right,

$b_1=3$, $b_2=3$, and for $n \ge 2$, $b_{n+1}b_{n-1}=b_n^2+2007$.

Find the largest integer less than or equal to $\dfrac{b_{2007}^2+b_{2006}^2}{b_{2007}b_{2006}}$.

We are given that

$b_{n+1}b_{n-1}=b_n^2+2007$--(1)

Replacing $n$ by $n-1$ we have

$b_{n}b_{n-2}=b_{n-1}^2+2007$

$b_{n-1}^2=b_{n}b_{n-2}-2007$--(2)

Adding the equations (1) and (2) gives

$b_{n-1}(b_{n+1}+b_{n-1})=b_{n}(b_{n}+b_{n-2})$

$\dfrac{b_{n+1}+b_{n-1}}{b_{n}}=\dfrac{b_{n}+b_{n-2}}{b_{n-1}}$

If we define $k_i=\dfrac{b_i}{b_{i-1}}$ for each $i \ge 2$, the equation above means

$k_{n+1}+\dfrac{1}{k_n}=k_n+\dfrac{1}{k_{n-1}}$

We can thus calculate that

$k_{2007}+\dfrac{1}{k_{2006}}=k_3+\dfrac{1}{k_{2}}=225$

Now, notice that $k_{2007}=\dfrac{b_{2007}}{b_{2006}}=\dfrac{b_{2006}^2+2007}{b_{2005}b_{2006}}>\dfrac{b_{2006}}{b_{2005}}=k_{2006}$.

This means that

$k_{2007}+\dfrac{1}{k_{2007}}<k_{2007}+\dfrac{1}{k_{2006}}=225$

It is only a tiny bit less because all of the $k_i$ are greater than 1, so we conclude that the floor of $\dfrac{b_{2007}^2+b_{2006}^2}{b_{2007}b_{2006}}=k_{2007}+\dfrac{1}{k_{2007}}$ is 224.