Question

Se considera numărul a n=5/1•2+9/2•3+15/3•4+. +n la puterea 2-n+3/(n-1)n,n€N,n>1.
Sa se determine[an]si{an}. ​.

Answer 1

Explicație pas cu pas:

n > 1

$T_{n - 1} = \dfrac{ {n}^{2} - n + 3 }{(n - 1) \cdot n} = \dfrac{ n \cdot (n - 1) + 3 }{(n - 1) \cdot n} = 1 + 3 \cdot \dfrac{1}{(n - 1) \cdot n}$

$= 1 + 3 \cdot \bigg(\dfrac{1}{n - 1} - \dfrac{1}{n} \bigg)$

atunci:

$a_{n} = \dfrac{5}{1 \cdot 2} + \dfrac{9}{2 \cdot 3} + \dfrac{15}{3 \cdot 4} + ... + \dfrac{ {n}^{2} - n + 3 }{(n - 1) \cdot n}$

$= 1 + 3 \cdot \bigg(\dfrac{1}{1} - \dfrac{1}{2} \bigg) + 1 + 3 \cdot \bigg(\dfrac{1}{2} - \dfrac{1}{3} \bigg) + 1 + 3 \cdot \bigg(\dfrac{1}{3} - \dfrac{1}{4} \bigg) + ... + 1 + 3 \cdot \bigg(\dfrac{1}{n - 1} - \dfrac{1}{n} \bigg)$

$= (n - 1) + 3 \cdot \bigg(\dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{n - 1} - \dfrac{1}{n} \bigg)$

$= (n - 1) + 3 \cdot \bigg(\dfrac{1}{1} - \dfrac{1}{n} \bigg) = (n - 1) + \dfrac{3(n - 1)}{n}$

$= \dfrac{(n + 3)(n - 1)}{n}$