Răspuns:
T 81
Explicație pas cu pas:
[tex] \boxed{ T_{k+1} = C_{n}^{k}\cdot a^{n-k}\cdot b^{k}}[/tex]
[tex]{\Big(\sqrt[3]{x} + \frac{2}{ \sqrt{x}} \Big)}^{200} = {\Big( {x}^{ \frac{1}{3}} + 2 \cdot {x}^{ - \frac{1}{2} } \Big)}^{200} \\ [/tex]
[tex]T_{k+1} = C_{200}^{k}\cdot {\Big( {x}^{ \frac{1}{3}}\Big)}^{200 - k} \cdot {\Big(2 \cdot {x}^{ - \frac{1}{2} } \Big)}^{k} = \\ = C_{200}^{k}\cdot {x}^{ \frac{200 - k}{3}} \cdot {2}^{k} \cdot {x}^{ - \frac{k}{2} } = C_{200}^{k}\cdot {2}^{k} \cdot {x}^{ \frac{200 - k}{3} - \frac{k}{2} } \\ = C_{200}^{k}\cdot {2}^{k} \cdot {x}^{ \frac{400 - 5k}{6}}[/tex]
[tex]{x}^{ \frac{400 - 5k}{6}} = {x}^{0} \implies \frac{400 - 5k}{6} = 0 \\ 400 - 5k = 0 \implies \bf k = 80[/tex]
=>
[tex]T_{80+1} = C_{200}^{80}\cdot {2}^{80} \cdot {x}^{0} \iff \bf T_{81} = C_{200}^{80}\cdot {2}^{80} \\ [/tex]
