Question

(2x - y)^4

Explanation:

Step1: 使用二项式定理\((a + b)^n=\sum_{k = 0}^{n}C_{n}^{k}a^{n - k}b^{k}\)，这里\(a = 2x\)，\(b=-y\)，\(n = 4\)

根据组合数公式\(C_{n}^{k}=\frac{n!}{k!(n - k)!}\)，分别计算\(k = 0,1,2,3,4\)时的项。

Step2: 当\(k = 0\)时

\(C_{4}^{0}(2x)^{4}(-y)^{0}=\frac{4!}{0!4!}\times16x^{4}\times1 = 16x^{4}\)

Step3: 当\(k = 1\)时

\(C_{4}^{1}(2x)^{3}(-y)^{1}=\frac{4!}{1!3!}\times8x^{3}\times(-y)=4\times8x^{3}\times(-y)=-32x^{3}y\)

Step4: 当\(k = 2\)时

\(C_{4}^{2}(2x)^{2}(-y)^{2}=\frac{4!}{2!2!}\times4x^{2}\times y^{2}=6\times4x^{2}y^{2}=24x^{2}y^{2}\)

Step5: 当\(k = 3\)时

\(C_{4}^{3}(2x)^{1}(-y)^{3}=\frac{4!}{3!1!}\times2x\times(-y^{3})=4\times2x\times(-y^{3})=-8xy^{3}\)

Step6: 当\(k = 4\)时

\(C_{4}^{4}(2x)^{0}(-y)^{4}=\frac{4!}{4!0!}\times1\times y^{4}=y^{4}\)

Step7: 合并所有项

将上述各项相加：\(16x^{4}-32x^{3}y + 24x^{2}y^{2}-8xy^{3}+y^{4}\)

Answer:

\(16x^{4}-32x^{3}y + 24x^{2}y^{2}-8xy^{3}+y^{4}\)