Question

O 10° termo da expressão $[27x^{5} +\frac{1}{3x} ]^{12}$

Answer 1

Resposta:

$\displaystyle T_{10}=220x^6$

Explicação passo-a-passo:

Binômio de Newton

$\displaystyle (A+B)^n\\\\T_{p+1}={n \choose p}A^{(n-p)}.B^{(p)}$

$\displaystyle (27x^5+\frac{1}{3x} )^{12} \\\\\\A=27x^5\\\\B=\frac{1}{3x}\\\\n=12\\\\T_{10}=?\\\\T_{p+1}=T_{10} \Rightarrow p+1=10\Rightarrow p=9$

$\displaystyle T_{10}={12 \choose 9}A^{(12-9)}.B^{(9)}=\frac{12!}{9!(12-9)!}A^3B^9=\frac{12.11.10.\diagup\!\!\!\!9!}{\diagup\!\!\!\!9!3.2.1}A^3B^9= 220A^3B^9\\\\\\T_{10}=220(27x^5)^3.(\frac{1}{3x})^9 =220.\diagup\!\!\!\!\!\!27^3x^{15}.(\frac{1}{\diagup\!\!\!\!\!\!3^9x^9})=220.x^{(15-9)}=220x^6$