Question

sa se determine termenul al 9 al dezvoltarii (2x^2+2/x)^12

Answer 1

$\displaystyle \mathtt{\left(2x^2+ \frac{2}{x}\right)^{12} }\\ \\ \mathtt{T_{9}=T_{8+1}=C_{12}^8\left(2x^2\right)^{12-8}\left( \frac{2}{x}\right)^8= \frac{12!}{(12-8)!\cdot8!}\cdot\left(2x^2\right)^4\cdot \frac{2^8}{x^8}=}\\ \\ \mathtt{= \frac{12!}{4!\cdot8!}\cdot2^4x^8\cdot \frac{2^8}{x^8}= \frac{8!\cdot9\cdot10\cdot11\cdot12}{1\cdot2\cdot3\cdot4\cdot8!}\cdot2^4\cdot2^8 =}\\ \\ \mathtt{=495\cdot16\cdot256=2027520\Rightarrow T_9=2027520}$