二项式定理例题
\[利用二项式列出(2+x)^{3}展开式中的各项 \]
\[\\ \\
\]
\[T_{1}=C_{0}^{3}\cdot 2^{3}x^{0}=1\cdot 2^{3}\cdot 1=8
\\
T_{2}=C_{1}^{3}\cdot2^{3-1}x^{1}=\frac{3!}{1!(3-1)!}\cdot 4x=12x
\\
T_{3}=C_{2}^{3}\cdot 2^{3-2} x^{2}=\frac{3!}{2!(3-2)!}\cdot 2x^{2}=6x^{2}
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T_{4}=C_{3}^{3}\cdot 2^{3-3}x^{3}=\frac{3!}{3!(3-3)!}\cdot x^{3}=x^{3}
\\
T_{1}+T_{2}+T_{3}+T_{4}=8+12x+6x^{2}+x^{3}
\]
\[求(x^{2}-\frac{1}{2x})的x^{9}项的系数
\]
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\]
\[(x^{2}-\frac{1}{2x}) \Rightarrow[x^{2}+(-\frac{1}{2}\cdot x^{-1})]^{9}
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T_{r+1}=C\tbinom{r}{9}\cdot (x^{2})^{9-r}\cdot (x{-1})^{r}\cdot
(-\frac{1}{2})^{r}
\\
T_{r+1}=C\tbinom{r}{9}x^{18-2r-r}\cdot (-\frac{1}{2})^{r}
\\
\because 18-3r=9,则r=3
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\therefore
C\tbinom{3}{9} (-\frac{1}{2})^{3}=\frac{9!}{3!(9-3)!}\cdot -\frac{1}{8}
\\
=-\frac{21}{2}
\]
\[求(1+2x)^{3}(1-x)^{4}的x^{2}项的系数\]
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\]
\[两个方程式的通项分别为:\\
(1+2x)^{3}=C\tbinom{m}{3}1^{3-m}(2x)^{m}= C\tbinom{m}{3}2^{m}x^{m},\\
m\in(0,1,2,3)
\]
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\]
\[(1-x)^{4}=C\tbinom{n}{4}1^{4-n}x^{n}=C\tbinom{n}{4}x^{n}(-1)^{n},\\
n\in(0,1,2,3,4)
\]
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\]
\[\because求的是x^{2}项,\therefore m+n=2,有3种情况满足m+n=2:\\
m=2,n=0;\\
m=n=1;\\
m=0,n=2;
\]
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\]
\[将3种情况一一带入并按照题目来计算:\\
C\tbinom{2}{3} 2^{2}x^{2}\cdot C\tbinom{0}{4}x^{0}(-1)^{0}+\\
C\tbinom{1}{3} 2^{1}x^{1}\cdot C\tbinom{1}{4}x^{1}(-1)^{1}+\\
C\tbinom{0}{3} 2^{0}x^{0}\cdot C\tbinom{2}{4}x^{2}(-1)^{2}=-6x^{2}
\]
