For $\alpha, \beta$ the roots of $x^2-x-1=0$, define $a_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}$ and $b_n=a_{n-1}+a_{n+1}$. Then ...

Question

Let $\alpha$ and $\beta$ be the roots of $x^{2}-x-1=0$, with $\alpha>\beta$. For all positive integers $n$, define $a_{n}=\frac{a^{n}-\beta^{n}}{\alpha-\beta}, \quad n \geq 1, b_{1}=1$ and $b_{n}=a_{n-1}+a_{n+1}, n \geq 2$. Then which of the following options is/are correct?

(A) $\sum_{n=1}^{\infty} \frac{b_{n}}{10^{n}}=\frac{8}{89}$

(B) $b_{n}=\alpha^{n}+\beta^{n}$ for all $n \geq 1$

(C) $a_{1}+a_{2}+a_{3}+\ldots .+a_{n}=a_{n+2}-1$ for all $n \geq 1$

(D) $\sum_{n=1}^{\infty} \frac{a_{n}}{10^{n}}=\frac{10}{89}$

I tried using basic algebraic manipulations and getting certain recurrence relations and proceeding. But the irony is this question was asked in an exam with max time 3 minutes per question. And I can't get anywhere even less than 15 minutes.

Any intelligent techniques will be appreciated which might not have clicked to me.

Answer 1

A first remark is that, recognizing Binet's formula, the $a_n$ can be identified with Fibonacci numbers $F_n$.

Using generating functions/series is quite natural in such a context.

• The $a_n=F_n$ have the classical generating function:

$$F(x):=\sum_{k=1}^{\infty}F_kx^k=\dfrac{x}{1-x-x^2}=\underbrace{1}_{F_1}x+\underbrace{1}_{F_2}x^2+\underbrace{2}_{F_3}x^3+\underbrace{3}_{F_4}x^4+...\tag{1}$$

(We can consider by assumption that $F_0=0$...)

Setting $x=\dfrac{1}{10}$ in (1) validates affirmation (D).

• The $b_n=F_{n-1}+F_{n+1}$ called Lucas numbers $L_n$, have the following generating function:

$$\sum_{k=2}^{\infty}(F_{k-1}+F_{k+1})x^k =\left(\frac{1}{x}+x\right)F(x)-\color{red}{1}=\dfrac{x+2x^2}{1-x-x^2}, \tag{2}$$

whose Taylor expansion is:

$$\sum_{k=1}^{\infty}L_{k}x^k =x+3x^2+4x^3+7x^4+11x^5+21x^6+\cdots$$

setting $x=\dfrac{1}{10}$ gives $\dfrac{12}{89}$ (as found by Ivan Kaznacheyeu using other means) invalidating affirmation (A). [NB: the presence of correcting term $\color{red}{1}$ is understood when writing the very first terms of (1)].

Final remark: Indeed, it looks strange that these questions are asked in the framework of a very limited range of time. The generating formula has to be known... Morever, what about less known Lucas numbers ?

Answer 2

$a_n=\frac1{\alpha-\beta}(\alpha^n-\beta^n)$

$\alpha^{n+2}=\alpha^n\alpha^2=$ $\alpha^n(\alpha+1)=\alpha^{n+1}+\alpha^n$

$a_{n+2}=a_{n+1}+a_n$

$\sum_{i=1}^n a_i=a_{n+2}-1$ can be easily shown by induction.

$\frac{a_n}{10^n}=\frac1{\alpha-\beta}\left((\frac{\alpha}{10})^n-(\frac{\beta}{10})^n\right)$

$\sum_{n=1}^\infty \frac{a_n}{10^n}=\frac1{\alpha-\beta}\left(\frac{\alpha}{10}\frac{1}{1-\frac{\alpha}{10}}-\frac{\beta}{10}\frac{1}{1-\frac{\beta}{10}}\right)=$ $\frac1{\alpha-\beta}\left(\frac{\alpha}{10-\alpha}-\frac{\beta}{10-\beta}\right)=\frac{10}{100-10(\alpha+\beta)+\alpha\beta}=\frac{10}{100-10+1}=\frac{10}{89}$

$\alpha^n+\beta^n=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}-\alpha\beta\frac{\alpha^{n-1}-\beta^{n-1}}{\alpha-\beta}=a_{n+1}+a_{n-1}=b_n$, $n\geq 2$

$b_1=1=\alpha+\beta$

$\frac{b_n}{10^n}=\left(\frac{\alpha}{10}\right)^n+\left(\frac{\beta}{10}\right)^n$

$\sum_{n=1}^\infty \frac{b_n}{10^n}=\frac{\alpha}{10}\frac{1}{1-\frac{\alpha}{10}}+\frac{\beta}{10}\frac{1}{1-\frac{\beta}{10}}=$ $\frac{\alpha}{10-\alpha}+\frac{\beta}{10-\beta}=$ $\frac{10(\alpha+\beta)-2\alpha\beta}{100-10(\alpha+\beta)+\alpha\beta}=\frac{12}{89}$.