Question

sa se determine al treilea termen din dezvoltarea:
$( \frac{ \sqrt[3]{a} }{\sqrt{b} } + \sqrt{a \times b} )^{6}$
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Answer 1

Răspuns:

Explicație pas cu pas:

$T_{k+1}=C_{n}^{k}a^{n-k}b^{k},~unde~k+1=3,~adica~k=2,~n=6,~a=\frac{\sqrt[3]{a} }{\sqrt{b}},~b=\sqrt{ab} ,~deci\\ T_{2+1}=C_{6}^{2}*(\frac{\sqrt[3]{a} }{\sqrt{b}})^{6-2}*(\sqrt{ab})^{2}=\frac{6!}{2!*4!} *\frac{a\sqrt[3]{a} }{b^{2}} *ab=\frac{5*6}{2}*\frac{a^{2}\sqrt[3]{a} }{b} =15*\frac{a^{2}\sqrt[3]{a} }{b}$