二项式系数练习题目
二项式系数练习题目
\(\mathcal{No.}1\)
\(3^{5}\times2^{13}\times\binom{18}{5}\)
\(0(无此项)\)
\(\mathcal{No.}2\)
\(\because \sum_{k=0}^n\binom{n}{k}x^{n-k}y^{k}=(x+y)^n\)
\(令 x=1,y=2\)
\(\therefore \sum_{k=0}^n\binom{n}{k}2^{k}=3^n\)
\(令 x=1,y=r\)
\(\therefore \sum_{k=0}^n\binom{n}{k}r^{k}=(r+1)^n\)
\(\mathcal{No.}3\)
\(\because \sum_{k=0}^n\binom{n}{k}x^{n-k}y^{k}=(x+y)^n\)
\(令 x=-1,y=3\)
\(\therefore \sum_{k=0}^n\binom{n}{k}3^{n-k}(-1)^{k}=(2)^n\)
\(\mathcal{No.}4\)
\(\because \sum_{k=0}^n\binom{n}{k}x^{n-k}y^{k}=(x+y)^n\)
\(令 x=-1,y=10\)
\(\therefore \sum_{k=0}^n\binom{n}{k}10^{k}(-1)^{n-k}=9^n\)
\(\therefore \sum_{k=0}^n\binom{n}{k}10^{k}(-1)^{k}=9^n\times(-1)^n\)
\(\mathcal{No.}5\)
\(即证 \binom{n}{k}=\binom{n-1}{k-1}+\binom{n-2}{k-1}+\binom{n-3}{k-1}+\binom{n-3}{k}\)
\(\because \binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}\)
\(\therefore \binom{n}{k}=\binom{n-1}{k-1}+\binom{n-2}{k-1}+\binom{n-2}{k}\)
\(\therefore \binom{n}{k}=\binom{n-1}{k-1}+\binom{n-2}{k-1}+\binom{n-3}{k-1}+\binom{n-3}{k}\)
\(\therefore \binom{n}{k}-\binom{n-3}{k}=\binom{n-1}{k-1}+\binom{n-2}{k-1}+\binom{n-3}{k-1}\)
\(\mathcal{No.}6\)
\(若为奇数\)
\(易证\)
\(若为偶数\)
\(令2m=n\)
\(\because(x+1)^n(x-1)^n=(x^2-1)^n\)
\(\therefore \sum\limits_{k=0}^n\binom{n}{k}x^k\times\sum\limits_{k=0}^n(-1)^k\binom{n}{k}x^k=\sum\limits_{k=0}^n\binom{n}{k}(-1)^kx^{2k}\)
\(又\because 第n项系数一定\)
\(\therefore \sum\limits_{k=0}^n\binom{n}{k}\binom{n}{n-k}(-1)^kx^{k}x^{n-k}=\binom{2m}{m}x^n(-1)^m\)
\(\therefore x^n\sum\limits_{k=0}^n{\binom{n}{k}}^2(-1)^k=\binom{2m}{m}x^n(-1)^m\)
\(\therefore \sum\limits_{k=0}^n{\binom{n}{k}}^2(-1)^k=\binom{2m}{m}(-1)^m\)
\(\mathcal{No.}7\)
\(\large{\binom{n+3}{k}}\)
\(考虑组合意义，原式即为钦定3个为额外元素，实际上原式的4项分别为额外元素选0,1,2,3的方案数\)
\(\mathcal{No.}8\)
\(\binom{r}{k}=\binom{r}{r-k}=\dfrac{r}{r-k}\times\binom{r-1}{r-k-1}=\dfrac{r}{r-k}\times\binom{r-1}{k}(???r好像为实数。。。。)\)
\(\mathcal{No.}9\)
\(原式=\sum_{k=0}^n\dfrac{\binom{n}{k}}{k+1}\times(-1)^k=\sum_{k=0}^n\dfrac{\binom{n+1}{k+1}}{n+1}\times(-1)^k=\dfrac{1}{n+1}\sum_{k=0}^n\binom{n+1}{k+1}\times(-1)^k=\dfrac{1}{n+1}\sum_{k=1}^{n+1}\binom{n+1}{k}\times(-1)^{k+1}=\dfrac{1}{n+1}\)
\(\mathcal{No.}10\)
\(即证 \binom{n+1}{k+1}=\sum_{i=0}^n\binom{i}{k}\)
\(考虑组合意义，即在前i+1个球中，必选i+1个球的方案数之和，即在n+1个球中选k+1个球\)
\(\mathcal{No.}11\)
\(考虑问题为在m范围内选择(x,y,z)三元组\)
\(m^3=\binom{m}{3}\times6+\binom{m}{2}\times3\times2+\binom{m}{1}\)
\(\therefore a=6,b=6,c=1\)
\(\mathcal{No.}12\)
\(\sum\limits_{k=1}^n\binom{n}{k}\binom{n}{k-1}=\sum\limits_{k=1}^n\binom{n}{n-k}\binom{n}{k-1} \tag*{(1)}\)
\(\dfrac{1}{2}\binom{2n+2}{n+1}-\binom{2n}{n}=\dfrac{1}{2}\binom{2n+1}{n}\times\dfrac{2n+2}{n+1}-\binom{2n}{n}=\binom{2n+1}{n}-\binom{2n}{n}=\binom{2n}{n-1}\tag*{(2)}\)
\((1)式中相当于把2n个元素分成两组，然后一组由n-k个，一组k-1个，一共\binom{2n}{n-1},即(2)\)
\(\mathcal{No.}13\)
\(\sum\limits_{k=1}^nk{\binom{n}{k}}^2=\sum\limits_{k=1}^nk{\binom{n}{k}}\binom{n}{k}=n\sum\limits_{k=1}^n{\binom{n}{k}}\binom{n-1}{k-1}\)
\(即证\sum\limits_{k=1}^n{\binom{n}{n-k}}\binom{n-1}{k-1}=\binom{2n-1}{n-1}\tag*{(1)}\)
\((1)式中相当于把2n-1个元素分成两组，然后一组由n-k个，一组k-1个，一共\binom{2n-1}{n-1},即为等式\)
