展开 (e^{x^2}-1)^4

解答

展开

(e^{x^{2}} - 1)^{4}

e^{4 x^{2}} - 4 e^{3 x^{2}} + 6 e^{2 x^{2}} - 4 e^{x^{2}} + 1

求解步骤

(e^{x^{2}} - 1)^{4}

使用二项式定理:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = e^{x^{2}}, b = - 1

= i = 0 \sum 4 (i 4) (e^{x^{2}})^{(4 - i)} (- 1)^{i}

展开求和

= \frac{4 !}{0 ! ( 4 - 0 ) !} (e^{x^{2}})^{4} (- 1)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (e^{x^{2}})^{3} (- 1)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (e^{x^{2}})^{2} (- 1)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (e^{x^{2}})^{1} (- 1)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (e^{x^{2}})^{0} (- 1)^{4}

化简

\frac{4 !}{0 ! ( 4 - 0 ) !} (e^{x^{2}})^{4} (- 1)^{0} : e^{4 x^{2}}

化简

\frac{4 !}{1 ! ( 4 - 1 ) !} (e^{x^{2}})^{3} (- 1)^{1} : - 4 e^{3 x^{2}}

化简

\frac{4 !}{2 ! ( 4 - 2 ) !} (e^{x^{2}})^{2} (- 1)^{2} : 6 e^{2 x^{2}}

化简

\frac{4 !}{3 ! ( 4 - 3 ) !} (e^{x^{2}})^{1} (- 1)^{3} : - 4 e^{x^{2}}

\frac{4 !}{4 ! ( 4 - 4 ) !} (e^{x^{2}})^{0} (- 1)^{4} = 1

= e^{4 x^{2}} - 4 e^{3 x^{2}} + 6 e^{2 x^{2}} - 4 e^{x^{2}} + 1

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