Question

7/(7*9)+7/(9*11)+7/(11*13)+⋯+7/(n-2)n, unde n nu este egal cu 0 şi nici cu 2

Answer 1

Salut,

Notăm cu S suma din membrul stâng.

[tex]\frac{7}{7\cdot 9}=\frac{7}{2}\cdot\left(\frac{1}{7}-\frac{1}{9}\right);\\\frac{7}{9\cdot 11}=\frac{7}{2}\cdot\left(\frac{1}{9}-\frac{1}{11}\right);\\\frac{7}{11\cdot 13}=\frac{7}{2}\cdot\left(\frac{1}{11}-\frac{1}{13}\right);\\\\\ldots\\\\\frac{7}{(n-4)\cdot (n-2)}=\frac{7}{2}\cdot\left(\frac{1}{n-4}-\frac{1}{n-2}\right);\\\frac{7}{(n-2)\cdot (n)}=\frac{7}{2}\cdot\left(\frac{1}{n-2}-\frac{1}{n}\right).[/tex]

Dacă adunăm membru cu membru toate aceste relaţii obţinem:

$S=\frac{7}{2}\cdot\left(\frac{1}{7}-\frac{1}{n}\right)=\frac{1004}{2015},\;sau\;\frac{1}{2}-\frac{7}{2n}=\frac{1004}{2015}\Rightarrow\frac{7}{2n}=\frac{1}{2}-\frac{1004}{2015}=\frac{2015-2008}{4030}=\frac{7}{4030},\;sau\;2n=4030,\;deci\;n=2015.$

Green eyes.