Question

Sa se determine :
a) termenul al treilea al dezvoltarii
$( x+ \frac{1}{x}) ^{9}$

b) termenul al patrulea al dezvoltarii
$(x^{2}+ \sqrt{x} ) ^{10}$

c) termenul din mijloc al dezvoltarii
$(x \sqrt{x} + \frac{1}{x \sqrt[3]{x} } ) ^{12}$

Answer 1

$\displaystyle \mathtt{a)~\left(x+ \frac{1}{x}\right)^9 }\\ \\ \mathtt{T_3=T_{2+1}=C_9^2x^{9-2}\left( \frac{1}{x}\right)^2= \frac{9!}{7! \cdot 2!}\cdot x^7\cdot \frac{1}{x^2} = \frac{7!\cdot8\cdot9}{7! \cdot 1 \cdot 2} \cdot x^7\cdot \frac{1}{x^2}= }\\ \\ \mathtt{=36x^7\cdot x^{-2}=36x^5\Rightarrow T_3=36x^5}$

$\displaystyle \mathtt{b)~\left(x^2+ \sqrt{x} \right)^{10}}\\ \\ \mathtt{T_4=T_{3+1}=C_{10}^3\left(x^2\right)^{10-3}\left( \sqrt{x} \right)^3= \frac{10!}{7! \cdot 3!} \cdot \left(x^2\right)^7\cdot \left(x^{ \frac{1}{2} \right)^3}=}\\ \\ \mathtt{= \frac{7! \cdot 8\cdot9\cdot10}{7!\cdot1\cdot2\cdot3} \cdot x^{14}\cdot x^{ \frac{3}{2} }=120x^{ \frac{31}{2} }\Rightarrow T_4=120x^{ \frac{31}{2} }}$

$\displaystyle \mathtt{c)~\left(x \sqrt{x} + \frac{1}{x \sqrt[\mathtt3]{\mathtt x} } \right)^{12}}\\ \\ \mathtt{T_7=T_{6+1}=C_{12}^6\left(x \sqrt{x} \right)^{12-6}\left( \frac{1}{x \sqrt[\mathtt3]{\mathtt x}}\right)^6 =\frac{12!}{6!\cdot6!}\cdot\left(x \cdot x^{ \frac{1}{2} }\right)^6\cdot\left( \frac{1}{x\cdot x^ {\frac{1}{3} }}\right)^6= }$
$\displaystyle \mathtt{= \frac{6!\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12}{6!\cdot1\cdot2\cdot3\cdot4\cdot5\cdot6}\cdot \left(x^{ \frac{3}{2}}\right)^6\cdot\left( \frac{1}{x^{ \frac{4}{3} }}\right)^6=924x^{ \frac{18}{2} }\cdot \left(x^{- \frac{4}{3} }\right)^6=}\\ \\ \mathtt{=924x^{9}\cdot x^{- \frac{24}{3} }=924x^9\cdot x^{-8}=924x\Rightarrow T_7=924x}$