Question

Sea

\begin{aligned} f(n) &=\frac{5+3\sqrt5}{10} \left(\frac{1+\sqrt5}{2}\right)^n\\ &\quad+\frac{5-3\sqrt5}{10} \left(\frac{1-\sqrt5}{2}\right)^n. \end{aligned}

Entonces

f(n+1)-f(n-1),

expresado en función de

f(n),

es igual a:

Let

\begin{aligned} f(n) &=\frac{5+3\sqrt5}{10} \left(\frac{1+\sqrt5}{2}\right)^n\\ &\quad+\frac{5-3\sqrt5}{10} \left(\frac{1-\sqrt5}{2}\right)^n. \end{aligned}

Then

f(n+1)-f(n-1),

expressed in terms of

f(n),

equals:

\dfrac12f(n)

f(n)

2f(n)+1

f^2(n)

\dfrac12\left(f^2(n)-1\right)

Llama

\alpha

y

\beta

a las dos bases exponenciales

Call the two exponential bases

\alpha

and

\beta

Ambas raíces de

r^2-r-1=0

satisfacen

r^{n+1}-r^{n-1}=r^n

Both roots of

r^2-r-1=0

satisfy

r^{n+1}-r^{n-1}=r^n

Cada base

r=\frac{1\pm\sqrt5}{2}

satisface

r^2=r+1,

por lo que

r^2-1=r.

Entonces

r^{n+1}-r^{n-1}=r^{n-1}(r^2-1)=r^n.

Al aplicar esta identidad término a término a la combinación lineal que define la función se obtiene

f(n+1)-f(n-1)=f(n).

Por lo tanto, la respuesta correcta es B.

Each base

r=\frac{1\pm\sqrt5}{2}

satisfies

r^2=r+1,

so

r^2-1=r.

Therefore

r^{n+1}-r^{n-1}=r^{n-1}(r^2-1)=r^n.

Applying this identity term by term to the defining linear combination gives

f(n+1)-f(n-1)=f(n).

Thus, the correct answer is B.