Matematické Fórum

veadet · 26. 01. 2019 01:48

a ako to urobime pre $n-1$ ?

laszky · 26. 01. 2019 01:56 — Editoval laszky (26. 01. 2019 02:00)

↑ veadet:

Ok, uz to urychlime...

Predpokladame, ze $D_{n-1}=-\sum_{i=1}^{n-1}\frac{1}{a_i}\prod_{j=1}^{n-1}a_j$ , pak

$D_n=a_nD_{n-1}-a_1a_2\dots a_{n-1} = a_n\left(-\sum_{i=1}^{n-1}\frac{1}{a_i}\prod_{j=1}^{n-1}a_j\right)-a_1a_2\dots a_{n-1} =$
$=-\sum_{i=1}^{n-1}\frac{1}{a_i}\prod_{j=1}^{n}a_j-\prod_{j=1}^{n-1}a_j = -\sum_{i=1}^{n-1}\frac{1}{a_i}\prod_{j=1}^{n}a_j-\frac{1}{a_n}\prod_{j=1}^{n}a_j = -\sum_{i=1}^{n}\frac{1}{a_i}\prod_{j=1}^{n}a_j$