Ars Mathematica: Solución 2 al problema 80

k_{i}

(k_{i} = \frac{1}{r_{i}})

(k_{1} + k_{2} + k_{3} + k_{4})^{2} = 2 (k_{1}^{2} + k_{2}^{2} + k_{3}^{2} + k_{4}^{2})

k_{4}

k_{1}, k_{2}

k_{3}

k_{4} = k_{1} + k_{2} + k_{3} \pm \sqrt{k_{1} k_{2} + k_{1} k_{3} + k_{2} k_{3}}

D_{n} = \frac{r}{n (n + 1)}

r_{n} = \frac{r}{2 n (n + 1)}

k_{n} = 2 n (n + 1)

r_{n}

r_{n - 1}, r

r

k_{n} = 2 n (n + 1)

k_{1} = 2 \cdot 2 = 4 \Rightarrow r_{1} = \frac{r}{4}

k_{n - 1} = 2 n (n - 1)

1 + 1 + 2 n (n - 1) \pm \sqrt{2 n (n - 1) + 2 n (n - 1) + 1} =

= 2 + 2 n^{2} - 2 n \pm 2 \sqrt{(2 n - 1)^{2}} = 2 + 2 n^{2} - 2 n \pm [4 n - 2] =

= {\begin{matrix} 2 + 2 n^{2} - 2 n + 4 n - 2 = 2 n^{2} + 4 n = 2 (n^{2} + 2 n) = 2 n (n + 1) \\ 2 + 2 n^{2} - 2 n - 4 n + 2 = 2 n^{2} - 6 n + 4 = 2 (n^{2} - 3 n + 2) = 2 (n - 1) (n - 2) \end{matrix} .

2 n (n - 1)

Ars Mathematica