Siruri.Demonstratie ca un numar este termen general.Done

Nikolle Dum Aug 26, 2012 11:08 pm

Cum se rezolva? Question

Fie $E(x)=x^{2}-4x+3,\bigtriangledown x\epsilon \mathbb{R}.$ . Pentru $n\epsilon \mathbb{N}, n\geqslant 4$ , definim sirul $(a_{n})_{n\geq 4}$ prin:
$a_{n}=\frac{1}{E_{(4)}}+\frac{1}{E_{(5)}}+\cdots +\frac{1}{E_{(n)}}.$
Aratati ca $a_{n}=\frac{(n-3)(3n-4)}{4(n-1)(n-2)},\bigtriangledown n\geq 4$

**Admin** Lun Aug 27, 2012 1:44 am

Se rezolva astfel:
$\begin{array}{l} E\left( k \right) = {k^2} - 4k + 3 = \left( {k - 3} \right)\left( {k - 1} \right) \Rightarrow \frac{1}{{E\left( k \right)}} = \frac{1}{{\left( {k - 3} \right)\left( {k - 1} \right)}} = \frac{1}{2}\left( {\frac{1}{{k - 3}} - \frac{1}{{k - 1}}} \right)\\ \frac{1}{{E\left( 4 \right)}} = \frac{1}{2} \times \left( {\frac{1}{{4 - 3}} - \frac{1}{{4 - 1}}} \right) = \frac{1}{2} \times \left( {1 - \frac{1}{3}} \right)\\ \frac{1}{{E\left( 5 \right)}} = \frac{1}{2} \times \left( {\frac{1}{{5 - 3}} - \frac{1}{{5 - 1}}} \right) = \frac{1}{2} \times \left( {\frac{1}{2} - \frac{1}{4}} \right)\\ \frac{1}{{E\left( 6 \right)}} = \frac{1}{2} \times \left( {\frac{1}{{6 - 3}} - \frac{1}{{6 - 1}}} \right) = \frac{1}{2} \times \left( {\frac{1}{3} - \frac{1}{5}} \right)\\ ......\\ \frac{1}{{E\left( n \right)}} = \frac{1}{2} \times \left( {\frac{1}{{n - 3}} - \frac{1}{{n - 1}}} \right) \end{array}$
Adunând acești termeni se reduc majoritatea .... ce ramne ?