Question

o termo em x10 no desenvolvimento de (x-1\x)20

Answer 1

Expressão:

$\displaystyle \left( x - \frac{1}{x} \right) ^{20}$

O enunciado pede que encontremos x¹⁰.

Basta usarmos o termo geral do Binômio de Newton:

$\displaystyle T_{p+1} = {n \choose p} \cdot a^{n-p} \cdot b^{p}$

Em que:

a = x
b = -1/x
n = expoente do binômio
p = valor que depende do termo T o qual queremos achar

Portanto:

$\displaystyle T_{9+1} = {20 \choose 9} \cdot x^{20-9} \cdot \left( \frac{-1}{x} \right)^{9}$

$\displaystyle T_{10} = C_{20,9} \cdot x^{11} \cdot \left( \frac{-1}{x} \right)^{9}$

$\displaystyle T_{10} = \frac{20!}{9! \cdot (20-9)!} \cdot x^{11} \cdot\left( \frac{-1}{x} \right)^9$

$\displaystyle T_{10} = \frac{20!}{9! \cdot 11!} \cdot x^{11} \cdot \left( \frac{-1}{x} \right)^9$

$\displaystyle T_{10} = \frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11!}{9! \cdot 11!} \cdot x^{11} \cdot \left( \frac{-1}{x} \right)^9$

$\displaystyle T_{10} = \frac{60.949.324.800}{9!} \cdot x^{11} \cdot\left( \frac{-1}{x} \right)^9$

$\displaystyle T_{10} = \frac{60.949.324.800}{ 362.880 } \cdot x^{11} \cdot \left( \frac{-1}{x} \right)^9$

$\displaystyle T_{10} = 167.960 \cdot x^{11} \cdot \left( \frac{-1}{x} \right)^9$

$\displaystyle T_{10} = 167.960 \cdot x^{11} \cdot \left[ \frac{1}{x} \cdot (-1) \right]^9$

$\displaystyle T_{10} = 167.960 \cdot x^{11} \cdot\left[ x^{-1} \cdot (-1) \right]^9$

$\displaystyle T_{10} = 167.960 \cdot x^{11} \cdot x^{-9} \cdot (-1)^9$

$\displaystyle T_{10} = 167.960 \cdot x^{11 - 9} \cdot (-1)$

$\displaystyle T_{10} = -167.960x^{2}$

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