Question

Se considera dezvoltarea (x^2 + 2/radical din x)^8 , x > 0. Sa se determine rangul termenului care il contine pe x^6.

Answer 1

$\it T_{k+1}=C_8^k (x^2)^{8-k}\cdot (\dfrac{2}{\sqrt{x}})^k\\ \\ \\ x^{16-2k}\cdot(x^{-\frac{1}{2}})^k =x^6 \Rightarrow \dfrac{32-4k-k}{2}=6 \Rightarrow 32-5k=12 \Rightarrow\\ \\ \Rightarrow5k=32-12 \Rightarrow5k=20 \Rightarrow k=4\\ \\ Deci,\ termenul\ al\ 5-lea\ con\c{\it t}ine\ pe\ x^6.$