Question

sa se determine termenul al 10-lea al dezvoltarii (a-1/a)^13

Answer 1

$T_{9+1}= C_{13}^9* a^{13-9}*(- \frac{1}{a})^9= -\frac{13!}{9!4!}*a^4* a^{-9}= \\ = -\frac{10*11*12*13}{1*2*3*4}* a^{-5}=- \frac{715}{a^5}$

Answer 2

Formula termenului general este:

$T_{k+1}=C_n^ka^{n-k}b^k$

În cazul nostru, $n=13,~a=a,~iar~b= -\dfrac{1}{a}$.

Al 10-lea termeni va fi:

[tex]T_{10}=T_{9+1}=C_{13}^9a^{13-9}\cdot(- \dfrac{1}{a} )^9\\\\ T_{10}=-C_{13}^9 a^4\cdot a^{-9} \\\\ T_{10}=- \dfrac{13!}{9!(13-9)!}\cdot a^{4-9}\\\\ T_{10}=- \dfrac{13!}{9!\cdot4!}\cdot a^{-5} \\\\ T_{10}=- \dfrac{9!\cdot10\cdot11\cdot12\cdot13}{9!\cdot1\cdot2\cdot3\cdot4} \cdot \dfrac{1}{a^5}[/tex]

Prin simplificări se ajunge la rezultatul:

$T_{10}=- \dfrac{715}{a^5}$