W rozwinięciu (3x²-[latex] frac{1}{x} [/latex])[latex]^{9} [/latex] znajdź stałe wyrażenie.

[latex](2x^2-frac{1}{x})^9=Sigma_{k=0}^9{9choose k}(3x^2)^{9-k}cdot(frac{1}{x})^k[/latex] [latex](x^2)^{9-k}cdot(x^{-1})^k=x^0\18-2k-k=0\18-3k=0\k=6[/latex] [latex]{9choose 6}cdot(3x^2)^3cdot(frac{1}{x})^6=frac{7cdot8cdot9}{2cdot3}cdot27x^6cdotfrac{1}{x^6}=7cdot4cdot3cdot27=2268[/latex]

n-k=3, k=6 [latex]\{9choose6}*(3x^2)^3*(x^{-1})^6= \ \ frac{9!}{3!*6!} *27x^6*x^{-6}=7*4*3*27=2268[/latex]