6.3 二项式定理
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知识剖析
二项式展开式
二项展开式的通项公式
二项式系数表(杨辉三角)
\((a+b)^{n}\)展开式的二项式系数,当\(n\)依次取\(1,2,3 …\)时,二项式系数表,表中每行两端都是\(1\),除\(1\)以外的每一个数都等于它肩上两个数的和.
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二项式系数的性质
(1)对称性:与首末两端“等距离”的两个二项式系数相等(\(\because C_{n}^{m}=C_{n}^{n-m}\)),直线\(r=\dfrac{n}{2}\)是图象的对称轴.
(2)增减性与最大值:
当\(n\)是偶数时,中间一项\(C_{n}^{\frac{n}{2}}\)取得最大值;当\(n\)是奇数时,中间两项 ,\(C_{n}^{\frac{n-1}{2}}, C_{n}^{\frac{n+1}{2}}\)取得最大值.
(3)二项式系数和:\(C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+\cdots+C_{n}^{r}+\cdots+C_{n}^{n}=2^{n}\),
奇数项的系数等于偶数项的系数等于\(2^{n-1}\),
\({\color{Red}{ 证明 }}\)
\(\because(1+x)^{n}=C_{n}^{0}+C_{n}^{1} x+\cdots+C_{n}^{r} x^{r}+\cdots+C_{n}^{n} x^{n}\)
令\(x=1\),则\(2^{n}=C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+\cdots+C_{n}^{r}+\cdots+C_{n}^{n}\),
令\(x=-1\),则\(C_{n}^{0}-C_{n}^{1}+C_{n}^{2}+\cdots+C_{n}^{r}+\cdots+(-1)^{n} C_{n}^{n}=0\),
奇数项的系数等于偶数项的系数等于\(2^{n-1}\).
\({\color{Red}{ 特别提醒}}\)
1.在运用二项式定理时一定要牢记通项公式\(T_{r+1}=C_{n}^{r} a^{n-r} b^{r}\) .另外二项展开式的二项式系数与该项的(字母)系数是两个不同的概念,前者只是指\(C_{n}^{r},\),而后者是指字母外的部分.
2.在使用通项公式\(T_{r+1}=C_{n}^{r} a^{n-r} b^{r}\)时,要注意通项公式是表示第\(r+1\)项,而不是第\(r\)项.
经典例题
【题型一】 二项式展开式
【典题1】若\(\left(x^{2}+\dfrac{1}{a x}\right)^{6}\)的展开式中,\(x^3\)的系数是\(-160\),则 ( )
A.\(a=-\dfrac{1}{2}\)
B.所有项系数之和为\(1\)
C.二项式系数之和为\(64\)
D.常数项为\(-320\)
【解析】由\(T_{r+1}=C_{6}^{r}\left(x^{2}\right)^{6-r}\left(\dfrac{1}{a x}\right)^{r}=\left(\dfrac{1}{a}\right)^{r} \cdot C_{6}^{r} \cdot x^{12-3 r}\),
令\(12-3r=3\),得\(r=3\).
\(\therefore \dfrac{1}{a^{3}} \cdot C_{6}^{3}=-160\),得\(a=-\dfrac{1}{2}\),故\(A\)正确;
\(\left(x^{2}+\dfrac{1}{a x}\right)^{6}=\left(x^{2}-\dfrac{2}{x}\right)^{6}\),
取\(x=1\),可得所有项系数之和为\(a_{0}+a_{1}+\cdots+a_{6}=(1-2)^{6}=1\),故\(B\)正确;
二项式系数之和为\(2^6=64\),故\(C\)正确;
\({\color{Red}{(二项式系数和: C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+\cdots+C_{n}^{r}+\cdots+C_{n}^{n}=2^{n}) }}\)
由\(12-3r=0\),得\(r=4\),展开式的常数项为\((-2)^{4} \cdot C_{6}^{4}=240\),故\(D\)错误.
\({\color{Red}{ (常数项即变量x的指数为0) }}\)
故选:\(ABC\).
【点拨】
① 先写出展开式的通项,并把其化为最简的形式;
② 每项的二项式系数\(C_n^r\)与其系数不是同一概念的.
【典题2】在二项式\((2 x+1)^6\)的展开式中,系数最大项的系数是( )
A.\(20\) \(\qquad \qquad \qquad \qquad\) B.\(160\) \(\qquad \qquad \qquad \qquad\) C.\(240\) \(\qquad \qquad \qquad \qquad\) D.\(192\)
【解析】二项式\((2 x+1)^6\)的展开式的通项为\(T_{k+1}=C_{6}^{k}(2 x)^{6-k}=2^{6-k} C_{6}^{k} x^{6-k}\),
设\(a_{k}=2^{6-k} C_{6}^{k}\),
则\(\dfrac{a_{k+1}}{a_{k}}=\dfrac{2^{5-k} C_{6}^{k+1}}{2^{6-k} C_{6}^{k}}=\dfrac{C_{6}^{k+1}}{2 C_{6}^{k}}=\dfrac{6-k}{2(k+1)}\)
当\(k≥2\)时,\(\dfrac{6-k}{2(k+1)}<1\),即\(\dfrac{a_{k+1}}{a_{k}}<1 \Rightarrow a_{k+1}<a_{k}\),即\(a_k\)递减;
而\(a_1<a_2\),故\(a_2\)取到最大值,
即系数最大项的系数为\(2^4 C_6^4=16× 15=240\).
【点拨】先求出系数通项,再利用求数列单调性的方法—作商法(作差法也行)求出最大项.
巩固练习
1(★★) [多选题]关于\(\left(x^{2}-\dfrac{2}{x}\right)^{5}\)的展开式,下列结论正确的是( )
A.奇数项的二项式系数和为\(32\)
B.所有项的系数和为\(-1\)
C.只有第\(3\)项的二项式系数最大
D.含\(x\)项的系数为\(-80\)
2(★★) [多选题]设常数\(a∈R\),\(n∈N^*\),对于二项式\((1+a \sqrt{x})^{n}\)的展开式,下列结论中,正确的是( )
A.若\(a<\dfrac{1}{n}\),则各项系数随着项数增加而减小
B.若各项系数随着项数增加而增大,则\(a>n\)
C.若\(a=-2\),\(n=10\),则第\(7\)项的系数最大
D.若\(a=-\sqrt{2}\),\(n=7\),则所有奇数项系数和为\(239\)
3(★★★) [多选题]设\((1+2 x)^{5}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}\),则满足\(a_{n}^{2}=2 a_{n-1} a_{n+1}\)的正整数\(n\)的值可能为( )
A.\(1\) \(\qquad \qquad \qquad \qquad\)B.\(2\) \(\qquad \qquad \qquad \qquad\) C.\(3\) \(\qquad \qquad \qquad \qquad\)D.\(4\)
4(★★★) 已知二项式\(\left(2 x+\dfrac{1}{\sqrt{x}}\right)^{n}\left(n \in N^{*}\right)\)的展开式中第\(2\)项与第\(3\)项的二项式系数之比是\(2:5\),按要求完成以下问题:
(1)求\(n\)的值;
(2)求展开式中常数项;
(3)计算式子\(C_{6}^{0} 2^{6}+C_{6}^{1} 2^{5}+C_{6}^{2} 2^{4}+C_{6}^{3} 2^{3}+C_{6}^{4} 2^{2}+C_{6}^{5} 2^{1}+C_{6}^{6} 2^{0}\)的值.
参考答案
- 【答案】\(BD\)
【解析】\(\left(x^{2}-\dfrac{2}{x}\right)^{5}\)的展开式的所有二项式系数和为\(32\),奇数项的二项式系数和为\(16\),故\(A\)错误;
取\(x=1\),可得所有项的系数和为\(-1\),故\(B\)正确;
\(\left(x^{2}-\dfrac{2}{x}\right)^{5}\)的展开式有\(6\)项,第3项与第四项的二项式系数相等且最大,故\(C\)错误;
展开式的通项为\(T_{r+1}=C_{5}^{r}\left(x^{2}\right)^{5-r}\left(-\dfrac{2}{x}\right)^{r}=(-2)^{r} C_{5}^{r} x^{10-3 r}\),
由\(10-3r=1\),得\(r=3\),
\(∴\)含\(x\)项的系数为\((-2)^3⋅C_5^3=-80\),故\(D\)正确.
故选:\(BD\). - 【答案】\(BCD\)
【解析】二项式\((1+a \sqrt{x})^{n}\)的展开式的通项为\(T_{r+1}=a^{r} C_n ^{r} x^{\frac{r}{2}}\),
对于\(A\):若\(a<0\),则各项系数一正一负交替出现,故\(A\)不对,
对于\(B\):\(C_{n}^{r} a^{r}<c_{n}^{r+1} a^{r+1}\)对于任意的\(r=0\),\(1\),\(2\),…,\(n-1\),都成立,
所以\(a>0\),且\(a>\dfrac{r+1}{n-r}\)对任意的\(r\)都成立,
\(∴a>n\),故\(B\)正确;
当\(a=-2\),\(n=10\),则展开式中奇数项的系数为正值,偶数项的系数为负值,
所以,只需比较\(C_{10}^{0}(-2)^{0}\),\(C_{10}^2 (-2)^2\),…,\(C_{10}^{6}(-2)^{6}\),\(C_{10}^{8}(-2)^{8}\),\(C_{10}^{10}(-2)^{10}\)即可,
可得,\(C_{10}^{6}(-2)^{6}\)最大,即展开式中第\(7\)项的系数最大,故\(C\)正确;
当\(a=-\sqrt{2}\),\(n=7\),
则奇数项系数和为:\(C_{7}^{0}(-\sqrt{2})^{0}+C_{7}^{2}(-\sqrt{2})^{2}+C_{7}^{4}(-\sqrt{2})^{4}+C_{7}^{6}(-\sqrt{2})^{6}=239\),故\(D\)正确;
故选:\(BCD\). - 【答案】\(BC\)
【解析】二项式的展开式的通项\(T_{n+1}=C_{5}^{n}(2 x)^{n}=C_{5}^{n} 2^{n} x^{n}\),
所以\(a_{n}=C_{5}^{n} 2^{n}\),要使\(a_{n}^{2}=2 a_{n-1} a_{n+1}\),
则\(\left(C_{5}^{n} 2^{n}\right)^{2}=2 C_{5}^{n-1} 2^{n-1} \cdot C_{5}^{n+1} 2^{n+1}\),
即\(\left(\dfrac{5 !}{n !(5-n) !}\right)^{2} \cdot 2^{2 n}=2 \times \dfrac{5 !}{(n-1) !(6-n) !} \times \dfrac{5 !}{(n+1) !(4-n) !} \times 2^{2 n}\),
化简得\(n^{2}-5 n+6=0\),解得\(n=2\)或\(3\),
故选:\(BC\). - 【答案】\((1)6\) \((2)60\) \((3)729\)
【解析】(1)二项式\(\left(2 x+\dfrac{1}{\sqrt{x}}\right)^{n}\left(n \in N^{*}\right)\)的展开式中第\(2\)项与第\(3\)项的二项式系数之比是 \(C_n^1:C_n^2=2:5\),
求得\(n=6\).
(2)展开式的通项公式为\(T_{r+1}=C_{6}^{r} \cdot 2^{6-r} \cdot x^{6-\frac{3 r}{2}}\),令\(6-\dfrac{3 r}{2}=0\),求得\(r=4\),
可得常数项为\(C_{6}^{4} \cdot 2^{2}=60\).
(3)\(C_{6}^{0} 2^{6}+C_{6}^{1} 2^{5}+C_{6}^{2} 2^{4}+C_{6}^{3} 2^{3}+C_{6}^{4} 2^{2}+C_{6}^{5} 2^{1}+C_{6}^{6} 2^{0}=(2+1)^{6}=3^{6}=729\).
【题型二】两个二项式相乘
【典题1】已知\(\left(1+\dfrac{a}{x^{2}}\right)\left(2 x-\dfrac{1}{x}\right)^{6}\)的展开式中各项系数的和为\(3\),则下列结论正确的有( )
A.\(a=2\)
B.展开式中常数项为\(64\)
C.展开式系数的绝对值的和\(2187\)
D.若\(r\)为偶数,则展开式中\(x^{r-2}\)系数是\(x^r\)系数的\(2\)倍
【解析】对于\(A\),
令\(x=1\),可得\(\left(1+\dfrac{a}{x^{2}}\right)\left(2 x-\dfrac{1}{x}\right)^{6}\)的展开式中各项系数的和为\((1+a)×1=3\),
\(∴a=2\),故\(A\)正确;
对于\(B\),易知\(\left(2 x-\dfrac{1}{x}\right)^{6}\)展开式中通项为\(T_{r+1}=C_{6}^{r}(2 x)^{6-r}\left(-\dfrac{1}{x}\right)^{r}=(-1)^{r} 2^{6-r} C_{6}^{r} x^{6-2 r}\)
其中\(r=\{0,1,2,3,4,5,6\}\),
即\(\left(2 x-\dfrac{1}{x}\right)^{6}=a_{0} x^{-6}+a_{1} x^{-4}+a_{2} x^{-2}+a_{3} x^{0}+a_{4} x^{2}+a_{5} x^{4}+a_{6} x^{6}\)
则\(\left(1+\dfrac{2}{x^{2}}\right)\left(2 x-\dfrac{1}{x}\right)^{6}=\left(1+\dfrac{2}{x^{2}}\right)\left(a_{0} x^{-6}+a_{1} x^{-4}+a_{2} x^{-2}+a_{3} x^{0}+a_{4} x^{2}+a_{5} x^{4}+a_{6} x^{6}\right)\),
则展开式中常数项为\(1\cdot a_3+2a_2\),
由\(T_{r+1}=(-1)^{r} 2^{6-r} C_{6}^{r} x^{6-2 r}\),易得\(a_2=48\),\(a_3=-160\),则\(a_3+2a_2=-64\),故\(B\)错误;
对于\(C\),
\(\left(1+\dfrac{2}{x^{2}}\right)\left(2 x-\dfrac{1}{x}\right)^{6}\)的展开式中各项系数绝对值的和,即项\(\left(1+\dfrac{2}{x^{2}}\right)\left(2 x+\dfrac{1}{x}\right)^{6}\)的各系数和,
令\(x=1\),为\((1+2)\cdot 3^6=2187\),故\(C\)正确;
对于\(D\)
由\(\left(1+\dfrac{2}{x^{2}}\right)\left(2 x-\dfrac{1}{x}\right)^{6}=\left(1+\dfrac{2}{x^{2}}\right)\left(a_{0} x^{-6}+a_{1} x^{-4}+a_{2} x^{-2}+a_{3} x^{0}+a_{4} x^{2}+a_{5} x^{4}+a_{6} x^{6}\right)\),
当\(r=-6\)时,\(x^{-6}\)的系数是\(a_{0}+2 a_{1}\),\(x^{-8}\)的系数是\(2a_0\),而\(a_1≠0\),故\(D\)不正确.
故选:\(AC\).
【点拨】对于二个二项式模型“多项式∙\((a+b)^{n}\)”,比如对于\(B\)选项,
想象下对\(\left(2 x-\dfrac{1}{x}\right)^{6}\)展开后的形式\(\left(1+\dfrac{2}{x^{2}}\right)\left(2 x-\dfrac{1}{x}\right)^{6}=\left(1+\dfrac{2}{x^{2}}\right)(\Delta+\Delta+\cdots+\Delta)\)
若要继续展开最后得到常数项,那只有\(1\)乘以\(\left(2 x-\dfrac{1}{x}\right)^{6}\)的常数项和\(2\)乘以\(\left(2 x-\dfrac{1}{x}\right)^{6}\)的\(x^{-2}\)项,
即所求的常数项\(=1\cdot a_3+2a_2\).
【典题2】\((1-\sqrt{x})^{6}(1+\sqrt{x})^{4}\)的展开式中\(x^2\)的系数为\(\underline{\quad \quad}\) .
【解析】\((1-\sqrt{x})^{6}(1+\sqrt{x})^{4}=[(1-\sqrt{x})(1+\sqrt{x})]^{4}(1-\sqrt{x})^{2}\)
\(=(1-x)^{4}(1-\sqrt{x})^{2}=(1-x)^{4}(1+x-2 \sqrt{x})\)
则\(x^2\)的项为\(1 \times C_{4}^{2}(-x)^{2}+x C_{4}^{1}(-x)=2 x^{2}\),
即\(x^2\)的系数为\(2\),
故选:\(B\).
【点拨】式子复杂,若能化简为熟悉的模型“多项式∙\((a+b)^{n}\)”,在求解过程中更便于思考.
巩固练习
1(★★) \(\left(x^{3}+6 x+1\right)\left(1-\dfrac{1}{x}\right)^{6}\)的展开式中的常数项为( )
A.\(-19\) \(\qquad \qquad \qquad \qquad\) B.\(-55\) \(\qquad \qquad \qquad \qquad\) C.\(21\)\(\qquad \qquad \qquad \qquad\).\(56\)
2(★★) 已知正整数\(n≥7\),若\(\left(x-\dfrac{1}{x}\right)(1-x)^{n}\)的展开式中不含\(x^4\)的项,则\(n\)的值为( )
A.\(7\) \(\qquad \qquad \qquad \qquad\) B.\(8\)\(\qquad \qquad \qquad \qquad\) C.\(9\) \(\qquad \qquad \qquad \qquad\) D.\(10\)
3(★★) \((1-x) \cdot\left(x+\dfrac{1}{x}+2\right)^{4}\)的展开式中\(x\)的系数是( )
A.\(10\) \(\qquad \qquad \qquad \qquad\) B.\(2\) \(\qquad \qquad \qquad \qquad\) C.\(-14\) \(\qquad \qquad \qquad \qquad\) D.\(34\)
4(★★★) \(\left(x+\dfrac{a}{x}\right)\left(2 x-\dfrac{1}{x}\right)^{5}\)的展开式中各项系数的和为\(2\),则其中正确命题的序号是( )
A.\(a=1\)
B.展开式中含\(x^6\)项的系数是\(-32\)
C.展开式中含\(x^{-1}\)项
D.展开式中常数项为\(40\)
参考答案
- 【答案】\(B\)
【解析】\(\left(x^{3}+6 x+1\right)\left(1-\dfrac{1}{x}\right)^{6}\)的展开式中的常数项为 \(C_{6}^{3} \cdot(-1)^{3}+6 \times C_{6}^{1} \times(-1)+C_{6}^{0}=-20-36+1=-55\),
故选:\(B\). - 【答案】\(B\)
【解析】正整数\(n≥7\),若\(\left(x-\dfrac{1}{x}\right)(1-x)^{n}\)的展开式中不含\(x^{4}\)的项,
则\((1-x)^{n}\)的展开式中的含\(x^{3}\)的项和含\(x^{5}\)的项的系数和为\(0\),
即\(-C_{n}^{3}+C_{n}^{5}=0\),\(∴n=8\),
故选:\(B\). - 【答案】\(C\)
【解析】\(\because(1-x) \cdot\left(x+\dfrac{1}{x}+2\right)^{4}=(1-x) \cdot\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^{8}\)
\(=(1-x) \cdot\left(C_{8}^{0} \cdot x^{4}+C_{8}^{1} \cdot x^{3}+C_{8}^{2} \cdot x^{2}+...+C_{8}^{8} \cdot \dfrac{1}{x^{4}}\right)\),
故展开式中\(x\)的系数是 \(C_8^3-C_8^4=-14\),
故选:\(C\). - 【答案】\(AD\)
【解析】令\(x=1\)则有\(1+a=2\),得\(a=1\),故二项式为\(\left(x+\dfrac{1}{x}\right)\left(2 x-\dfrac{1}{x}\right)^{5}\),
\(\left(2 x-\dfrac{1}{x}\right)^{5}\)通项公式为\((-1)^{r} 2^{5-r} C_{5}^{r} x^{5-2 r}\),\(r\)依次为\(0\),\(1\),\(2\),\(3\),\(4\),\(5\)
\(\left(x+\dfrac{1}{x}\right)\left(2 x-\dfrac{1}{x}\right)^{5}\)的展开式中含\(x^{6}\)项系数为\(\left(2 x-\dfrac{1}{x}\right)^{5}\)通项展开式式中\(x^{5}\)项系数的与\(x^{7}\)项的系数之和,
令\(5-2r=5\)解得\(r=0\),所以\(\left(2 x-\dfrac{1}{x}\right)^{5}\)通项展开式式\(x^{5}\)项系数\((-1)^{0} 2^{5} C_{5}^{0}=32\),
令\(5-2r=7\)解得\(r=-1\),不合题意,
\(∴\)展开式中含\(x^{6}\)项的系数是\(32\),
\(\left(x+\dfrac{1}{x}\right)\left(2 x-\dfrac{1}{x}\right)^{5}\)的展开式中含\(x^{-1}\)项系数为\(\left(2 x-\dfrac{1}{x}\right)^{5}\)通项展开式式中\(x^{-2}\)项系数的与常数项之和,
令\(5-2r=-2\),解得\(r=\dfrac{7}{2}\),不合题意,
令\(5-2r=0\),解得\(r=\dfrac{5}{2}\),不合题意,
则展开式不含\(x^{-1}\)项,
\(\left(x+\dfrac{1}{x}\right)\left(2 x-\dfrac{1}{x}\right)^{5}\)的展开式中含常数项为\(\left(2 x-\dfrac{1}{x}\right)^{5}\)通项展开式式中\(x^{-1}\)项系数的与\(x\)项的系数之和,
令\(5-2r=-1\),解得\(r=3\),令\(5-2r=1\),解得\(r=2\),
所以其常数项为\(-2^{2} \times C_{5}^{3}+2^{3} C_{5}^{2}=40\).
故选:\(AD\).
【题型三】 多项式展开式
【典题1】\(\left(x^{2}-4 x+\dfrac{1}{x}\right)^{5}\)的展开式中\(x^2\)项的系数为( )
A.\(840\) \(\qquad \qquad \qquad \qquad\)B.\(-600\) \(\qquad \qquad \qquad \qquad\) C.\(480\) \(\qquad \qquad \qquad \qquad\) D.\(-360\)
【解析】\(\left(x^{2}-4 x+\dfrac{1}{x}\right)^{5}=\left[x^{2}+\left(-4 x+\dfrac{1}{x}\right)\right]^{5}\),它展开式通项为\(T_{r+1}=C_{5}^{r} x^{10-2 r}\left(-4 x+\dfrac{1}{x}\right)^{r}\),
对于\(\left(-4 x+\dfrac{1}{x}\right)^{r}\),它展开式通项为\(C_{r}^{k}(-4)^{r-k} x^{r-2 k}\),其中\(r\)、\(k\)为非负整数且\(k≤r≤5\).
\({\color{Red}{(特别注意r 、k的限制范围) }}\)
多项式展开式中\(x\)的幂指数为\(10-2 r+r-2 k=10-r-2 k\),
求\(x^2\)的系数,则令\(10-r-2 k=2\),
可得\(\left\{\begin{array}{c}
k=0 \\
r=8>5
\end{array}\right.\)(舍去),\(\left\{\begin{array}{c}
k=1 \\
r=6>5
\end{array}\right.\)(舍去),\(\left\{\begin{array}{l}
k=2 \\
r=4
\end{array}\right.\),\(\left\{\begin{array}{c}
k=3 \\
r=2<k
\end{array}\right.\)(舍去),\(\left\{\begin{array}{c}
k=4 \\
r=0<k
\end{array}\right.\)(舍去),
\({\color{Red}{(利用r、k的限制范围排除某些结果) }}\)
所以只有\(\left\{\begin{array}{l}
k=2 \\
r=4
\end{array}\right.\)成立,
故展开式中\(x^2\)项的系数为 \(C_{5}^{4} \cdot C_{4}^{2} \cdot(-4)^{4-2}=480\),
故选:\(C\).
【点拨】① 多项式展开式,可转化为二项式展开式,本题把\(-4 x+\dfrac{1}{x}\)看成“一项”,其实也可以把“\(x^2-4x\)”看成一项;
② 本题利用了二次展开式,得到最后变量\(x\)的指数\(10-r-2k\),此时要特别注意\(r、k\)的限制范围.
巩固练习
1(★★) 在\(\left(1-x+\dfrac{1}{x^{2021}}\right)^{8}\)的展开式中,\(x^2\)的系数为( )
A.\(2021\) \(\qquad \qquad \qquad \qquad\) B.\(28\) \(\qquad \qquad \qquad \qquad\) C.\(-28\) \(\qquad \qquad \qquad \qquad\) D.\(-56\)
2(★★) \((x+y-z)^6\)的展开式中\(xy^2 z^3\)的系数是\(\underline{\quad \quad}\).
3(★★) 已知等差数列\(\left\{a_{n}\right\}\)的第\(5\)项是\(\left(x-\dfrac{1}{x}+2 y\right)^{6}\)展开式中的常数项,则\(a_2+a_8=\)\(\underline{\quad \quad}\).
参考答案
- 【答案】\(B\)
【解析】由于\(=\left(1-x+\dfrac{1}{x^{2021}}\right)^{8}\)的表示\(8\)个因式\(\left(1-x+\dfrac{1}{x^{2021}}\right)\)的乘积,
故有\(2\)个因式取\(x\),其余的因式都取\(1\),即可得到含\(x^{2}\)的项,
故\(x^{2}\)的系数\(C_8^2=28\),
故选:\(B\). - 【答案】\(-60\)
【解析】\((x+y-z)^{6}\)表示\(6\)个因式\((x+y-z)\)的乘积,故其中有一个因式取\(x\),其中\(2\)个因式取\(y\),其余的因式都取\(-z\),
即可得到展开式中\(x y^{2} z^{3}\)的项,
故该项的系数为\(C_{6}^{1} \cdot C_{5}^{2} \cdot C_{3}^{3} \cdot(-1)^{3}=-60\),
故答案为:\(-60\). - 【答案】\(-40\)
【解析】\(\because\left(x-\dfrac{1}{x}+2 y\right)^{6}\)表示\(6\)个因式\(\left(x-\dfrac{1}{x}+2 y\right)\)的乘积,
故当有\(3\)个因式取\(x\),其余的\(3\)个因式取\(-\dfrac{1}{x}\) 时,可得它的常数项为\(-C_{6}^{3} \cdot C_{3}^{3}=-20=a_{5}\),
等差数列\(\{a_n\}\)的第\(5\)项是\(\left(x-\dfrac{1}{x}+2 y\right)^{6}\)展开式中的常数项,则\(a_{2}+a_{8}=2 a_{5}=-40\)
【题型四】系数问题
【典题1】已知\(\left(1-x^{2}\right)(x+2)^{4}=a_{0}+a_{1}(x+1)+a_{2}(x+1)^{2}+\)\(a_{3}(x+1)^{3}+a_{4}(x+1)^{4}+a_{5}(x+1)^{5}+a_{6}(x+1)^{6}\),则( )
A.\(a_0=0\)\(\qquad \qquad\) B.\(a_3=20\) \(\qquad \qquad\)C.\(a_1+a_5=0\) \(\qquad \qquad\)D.\(a_2+a_4+a_6=a_1+a_3+a_5\)
【解析】对于\(A\),令\(x=-1\),则\(a_0=0\),\(A\)正确;
对于\(B\),
令\(t=x+1⇒x=t-1\),则已知等式变成
\(\left(2 t-t^{2}\right)(t+1)^{4}=a_{0}+a_{1} t+a_{2} t^{2}+a_{3} t^{3}+a_{4} t^{4}+a_{5} t^{5}+a_{6} t^{6}\),
\(∵(t+1)^4\)展开式通项为\(C_{4}^{r} t^{4-r}\),
\(\therefore a_{3}=2 C_{4}^{2}+(-1) C_{4}^{3}=8\).\(B\)错误;
对于\(C\),
令\(t=1\),得\(a_0+a_1+a_2+a_3+a_4+a_5+a_6=2^4=16\),
令\(t=-1\),得\(a_0-a_1+a_2-a_3+a_4-a_5+a_6=0\)
\(\therefore a_{1}+a_{3}+a_{5}=\dfrac{16}{2}=8\),\(∴a_1+a_5=0\),\(C\)正确;
对于\(D\),
令\(t=-1\),得\(a_0-a_1+a_2-a_3+a_4-a_5+a_6=0\)\(⇒a_0+a_2+a_4+a_6=a_1+a_3+a_5\)
又\(∵a_0=0\),\(∴a_2+a_4+a_6=a_1+a_3+a_5\),\(D\)正确,
故选:\(ACD\).
【点拨】① 对于类似系数问题,常令\(x=0\),\(x=1\),\(x=-1\)或根据等式结构取其他特殊值,这样往往能够得到展开式中某些系数的关系,这个要多尝试;
② 题目中等式右边(它是以\(x+1\)展开的),不是我们熟悉的按\(x\)来展开,那可以用换元法,把不熟悉的问题转化为熟悉的问题是求解数学题的常用思考模式.
【典题2】若\((1+x)+(1+x)^{2}+\cdots+(1+x)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}\),且\(a_{1}+a_{2}+\cdots+a_{n-1}=125-n\),则下列结论正确的是( )
A.\(n=6\)
B.\((1+2 x)^{n}\)展开式中二项式系数和为\(729\)
C.\((1+x)+(1+x)^{2}+\cdots+(1+x)^{n}\)展开式中所有项系数和为\(126\)
D.\(a_1+2a_2+3a_3+⋯+na_n=321\)
【解析】对于\(A\),
\(\because(1+x)+(1+x)^{2}+\cdots+(1+x)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}\),
\(∴\)较易得到\(a_0=n\),\(a_n=1\),
令\(x=1\),可得\(2+2^2+⋯+2^n=a_0+a_1+a_2+⋯+a_n=n+(125-n)+1=126\),
\(\Rightarrow \dfrac{2\left(1-2^{n}\right)}{1-2}=126 \Rightarrow n=6\),故\(A\)正确;
对于\(B\),
\((1+2 x)^{n}\)展开式中二项式系数和为\(2^{n}=2^{6}=64\),故\(B\)错误;
对于\(C\),
\((1+x)+(1+x)^{2}+\cdots+(1+x)^{n}\)展开式中所有项系数和\((1+x)+(1+x)^{2}+\cdots+(1+x)^{n}\),
故\(C\)正确;
对于\(D\),
\(\because(1+x)+(1+x)^{2}+\cdots+(1+x)^{6}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{6} x^{6}\),
两边求导得
\(1+2(1+x)+3(1+x)^{2}+\cdots+6(1+x)^{5}=a_{1}+2 a_{2} x+3 a_{3} x^{2}+\cdots+6 a_{6} x^{5}\),
令\(x=1\)得\(a_1+2a_2+3a_3+⋯+6a_6=1+4+12+32+80+192=321\),故\(D\)正确.
故选:\(ACD\).
【点拨】对于\(D\)选项,\(a_1+2a_2+3a_3+⋯+na_n\),每项的\(a_k\)前还有个系数,在原展开式中令\(x\)取什么值都无法得到这形式,对展开式两边取导数再给\(x\)取数是个巧妙的方法.
巩固练习
1(★★) [多选题]已知\((2+x)(1-2 x)^{5}=\)\(a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\),则( )
A.\(a_0\)的值为\(2\)
B.\(a_5\)的值为\(16\)
C.\(a_1+a_2+a_3+a_4+a_5+a_6\)的值为\(-5\)
D.\(a_1+a_3+a_5\)的值为\(120\)
2(★★★) [多选题]已知\((x-2)^{10}=a_{0}+a_{1}(x-1)+a_{2}(x-1)^{2}+\cdots+a_{10}(x-1)^{10}\),则下列结论正确的有( )
A.\(a_0=1\)
B.\(a_6=-210\)
C.\(\dfrac{a_{1}}{2}+\dfrac{a_{2}}{2^{2}}+\dfrac{a_{3}}{2^{3}}+\cdots+\dfrac{a_{10}}{2^{10}}=-\dfrac{1023}{1024}\)
D.\(a_{0}+a_{2}+a_{4}+a_{6}+a_{8}+a_{10}=512\)
3(★★★) [多选题]已知\((2 x-3)(x-2)^{8}=\)\(a_{0}+a_{1}(x-1)+a_{2}(x-1)^{2}+a_{3}(x-1)^{3}+\cdots+a_{9}(x-1)^{9}\),则下列结论正确的是( )
A.\(a_1+a_2+⋯+a_9=1\)
B.\(a_5=84\)
C.\(\dfrac{a_{1}}{2}+\dfrac{a_{2}}{2^{2}}+\cdots+\dfrac{a_{9}}{2^{9}}=1\)
D.\(a_1+2a_2+⋯+9a_9=0\)
参考答案
- 【答案】\(ABC\)
【解析】∵已知\((2+x)(1-2x)^5=a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+a_5 x^5+a_6 x^6\),
令等式中的\(x=0\),可得\(a_0=2\),故\(A\)正确.
\(a_{5}\)的值,即展开式中\(x^{5}\)的系数,为\(2×(-2)^5 C_5^5+(-2)^4 C_5^4=16\),故\(a_{5}=16\)正确.
在所给的等式中,令\(x=1\),得\(a_{0}+a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=-3\)①,又\(a_{0}=2\),
\(\therefore a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=-5\),故\(C\)正确;
在所给的等式中,令\(x=-1\),得\(a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}+a_{6}=243\)②,
由①②得:\(a_{1}+a_{3}+a_{5}=-123\),\(D\)错误.
故选:\(ABC\). - 【答案】\(ACD\)
【解析】\(\because(x-2)^{10}=a_{0}+a_{1}(x-1)+a_{2}(x-1)^{2}+\cdots+a_{10}(x-1)^{10}\),
令\(x=1\),得\(a_{0}=1\),故\(A\)正确,
令\(x=2\),得\(a_{0}+a_{1}+a_{2}+\cdots+a_{9}+a_{10}=0\),
令\(x=0\),得\(a_{0}-a_{1}+a_{2}+\cdots-a_{9}+a_{10}=2^{10}\),所以\(a_{0}+a_{2}+a_{4}+a_{6}+a_{8}+a_{10}=\dfrac{0+2^{10}}{2}=512\),故\(D\)正确;
令\(x=\dfrac{3}{2}\),得\(a_{0}+\dfrac{a_{1}}{2}+\dfrac{a_{2}}{2^{2}}+\cdots+\dfrac{a_{10}}{2^{10}}=\left(\dfrac{1}{2}\right)^{10}\),
所以\(\dfrac{a_{1}}{2}+\dfrac{a_{2}}{2^{2}}+\cdots+\dfrac{a_{10}}{2^{10}}=\dfrac{1}{2^{10}}-1=-\dfrac{1023}{1024}\),故\(C\)正确,
\(\because(x-2)^{10}=[(x-1)-1]^{10}=\)\(a_{0}+a_{1}(x-1)+a_{2}(x-1)^{2}+\cdots+a_{10}(x-1)^{10}\),
\(\therefore a_{6}=C_{10}^{6} \cdot(-1)^{4}=210\),故\(B\)错误,
故选:\(ACD\). - 【答案】\(ACD\)
【解析】\(\because(2 x-3)(x-2)^{8}=a_{0}+a_{1}(x-1)+a_{2}(x-1)^{2}+\)\(a_{3}(x-1)^{3}+\ldots+a_{9}(x-1)^{9}\),
令\(x=1\),得\(a_{0}=-1\),
令\(x=2\),得\(a_{0}+a_{1}+a_{2}+\ldots+a_{9}=0\),所以\(a_{1}+a_{2}+\ldots+a_{9}=1\),故\(A\)正确;
由\((2 x-3)(x-2)^{8}=[2(x-1)-1][(x-1)-1]^{8}\),
所以\(a_{5}=2 \times C_{8}^{4} \times(-1)^{4}-C_{8}^{3} \times(-1)^{3}=196\),故\(B\)错误;
令\(x=\dfrac{3}{2}\),
得\(\left(2 \times \dfrac{3}{2}-3\right)\left(\dfrac{3}{2}-2\right)^{8}=a_{0}+a_{1}\left(\dfrac{3}{2}-1\right)+a_{2}\left(\dfrac{3}{2}-1\right)^{2}+\)\(a_{3}\left(\dfrac{3}{2}-1\right)^{3}+\cdots+a_{9}\left(\dfrac{3}{2}-1\right)^{9}\),
所以\(a_{0}+\dfrac{a_{1}}{2}+\dfrac{a_{2}}{2^{2}}+\cdots+\dfrac{a_{9}}{2^{9}}=0\),
又\(a_{0}=-1\),所以\(\dfrac{a_{1}}{2}+\dfrac{a_{2}}{2^{2}}+\cdots+\dfrac{a_{9}}{2^{9}}=1\),故\(C\)正确;
设\(f(x)=(2 x-3)(x-2)^{8}=a_{0}+a_{1}(x-1)+a_{2}(x-1)^{2}+a_{3}(x-1)^{3}+\)\(\cdots+a_{9}(x-1)^{9}\),
则\(f^{\prime}(x)=2(x-2)^{8}+8(2 x-3)(x-2)^{7}=a_{1}+2 a_{2}(x-1)+\)\(3 a_{3}(x-1)^{2}+\cdots+9 a_{9}(x-1)^{8}\),
令\(x=2\),得\(=a_{1}+2 a_{2}+\ldots+9 a_{9}=0\),故\(D\)正确,
故选:\(ACD\).
【题型五】 其他应用
【典题1】证明\(3^{2 n+2}-8 n-9\)能被\(64\)整除\((n∈N^*)\).
【证明】\(3^{2 n+2}-8 n-9=9^{n+1}-8 n-9=(8+1)^{n+1}-8 n-9\)
\(\begin{aligned}
&=8^{n+1}+C_{n+1}^{1} \cdot 8^{n}+\cdots+C_{n+1}^{n-1} \cdot 8^{2}++C_{n+1}^{n} \cdot 8+1-8 n-9 \\
&=8^{n+1}+C_{n+1}^{1} \cdot 8^{n}+\cdots+C_{n+1}^{n-1} \cdot 8^{2}+8(n+1)+1-8 n-9 \\
&=8^{n+1}+C_{n+1}^{1} \cdot 8^{n}+\cdots+C_{n+1}^{n-1} \cdot 8^{2} \\
&=8^{2}\left(8^{n-1}+C_{n+1}^{1} 8^{n-2}+\cdots+C_{n+1}^{n-1}\right) \\
&=64\left(8^{n-1}+C_{n+1}^{1} 8^{n-2}+\cdots+C_{n+1}^{n-1}\right)
\end{aligned}\)
\(\because 8^{n-1}+C_{n+1}^{1} 8^{n-2}+\cdots+C_{n+1}^{n-1}\)是整数,
\(∴3^{2 n+2}-8 n-9\)能被\(64\)整除.
【点拨】这是整除与余数的问题,由于证明中的除数是\(64\),则要在\(3^{2 n+2}-8 n-9\)中尽量找到与其有关信息,没直接信息与\(64\)有关,而\(3^{2 n+2}=9^{n+1}=(8+1)^{n}\)中含有8的信息,这就找到了可靠的突破口,可往下演算尝试.
【典题2】求\(0.998^{6}\)的近似值,使误差小于\(0.001\).
【解析】\(0.998^{6}\)
\(\begin{aligned}
&=(1-0.002)^{6}=1+6 \cdot(-0.002)+15 \cdot(-0.002)^{2}+. .+(-0.002)^{6} \\
&\approx 1+6 \cdot(-0.002) \\
&=0.988
\end{aligned}\)
【点拨】这是求近似值,由于\(0.998\)接近\(1\),则由\(0.998^{6}=(1-0.002)^{6}\)进行演算.
【典题3】求证:\(C_{n}^{0}+3 C_{n}^{1}+5 C_{n}^{2}+\cdots+(2 n+1) C_{n}^{n}=(n+1) 2^{n}\).
【证明】设\(S_{n}=C_{n}^{0}+3 C_{n}^{1}+5 C_{n}^{2}+\cdots+(2 n+1) C_{n}^{n}\) ①
把①式右边倒转过来得
\(S_{n}=(2 n+1) C_{n}^{n}+(2 n-1) C_{n}^{n-1}+\cdots+3 C_{n}^{1}+C_{n}^{0}\),
又由\(C_{n}^{m}=C_{n}^{n-m}\)可得
\(S_{n}=(2 n+1) C_{n}^{0}+(2 n-1) C_{n}^{1}+\cdots+3 C_{n}^{n-1}+C_{n}^{n}\) ②
①+②得 \(2 S_{n}=(2 n+2)\left(C_{n}^{0}+C_{n}^{1}+\cdots+C_{n}^{n-1}+C_{n}^{n}\right)=2(n+1) \cdot 2^{n}\),
\(\therefore S_{n}=(n+1) \cdot 2^{n}\),
即\(C_{n}^{0}+3 C_{n}^{1}+5 C_{n}^{2}+\cdots+(2 n+1) C_{n}^{n}=(n+1) 2^{n}\),
原等式得证.
【点拨】这是证明“左式=右式”的题型,方法很多,
① 直接把左式化简得到右式,本题就是这样,它借鉴了数列中的“倒序相加法”,主要是留意到组合数的性质\(C_{n}^{m}=C_{n}^{n-m}\);
② 左式,右式同步化简,化简为同一结果,则左式=右式;
③ 数学归纳法对于与正整数\(n\)有关的等式或不等式均较为友好.
【典题4】用二项式定理证明:\(2^n>n^2 (n≥5)\).
【证明】\(∵n≥5\),\(∴n-2≥3\),
由二项式定理可得
\(2^{n-2}=(1+1)^{n-2}=C_{n-2}^{0}+C_{n-2}^{1}+C_{n-2}^{2}+\cdots+C_{n-2}^{n-2}\)
\(=1+(n-2)+\dfrac{(n-2)(n-3)}{2}+\cdots+C_{n-2}^{n-2}\)
\(>n-1+\dfrac{(n-2)(n-3)}{2}=\dfrac{n^{2}}{2}-\dfrac{3 n}{2}+2\).
\(\because 2^{n}-n^{2}=4 \cdot 2^{n-2}-n^{2}\)\(>4\left[\dfrac{n^{2}}{2}-\dfrac{3 n}{2}+2\right]-n^{2}=n^{2}-6 n+8=(n-3)^{2}-1\).
当\(n≥5\)时,\((n-3)^{2}-1>3>0\),
\(∴n≥5\)时,\(2^n>n^2\).
【点拨】不等式的证明常用的方法有放缩法,而二项式的展开式是放缩法中的一种方式,展开式中有多项,那可有选择的把“影响大的项”留下,去除“影响小的项”,从而达到放缩的目的,留“几项”就看放缩的要求了,在求近似值也是类似的方法.
巩固练习
1(★★) 若\(n\)是正奇数,则\(7^{n}+C_{n}^{1} 7^{n-1}+C_{n}^{2} 7^{n-2}+\cdots \ldots+C_{n}^{n-1} 7\)被\(9\)除的余数为( )
A.\(2\) \(\qquad \qquad \qquad \qquad\)B.\(5\) \(\qquad \qquad \qquad \qquad\) C.\(7\) \(\qquad \qquad \qquad \qquad\)D.\(8\)
2(★★) 用二项式定理证明:\(11^{10}-1\)能被\(100\)整除.
3(★★) 求\(1.02^{8}\)的近似值(精确到小数点后三位).
4(★★) 求和\(W=C_{n}^{0}+4 C_{n}^{1}+7 C_{n}^{2}+10 C_{n}^{3}+\cdots+(3 n+1) C_{n}^{n}\).
5(★★) 用二项式定理证明:\(\left(1+\dfrac{1}{k+1}\right)^{k+1} \geq 2\).
6(★★★) 记\(f(a)\)为\((a x+1)^{n-}\)二项展开式中的\(x^3\)项的系数,其中\(a∈\{1,2,3,…,n\}\),\(n≥3\).
(1)求\(f(1)\),\(f(2)\),\(f(3)\)
(2)证明:\(\sum_{a=1}^{n} f(a)=C_{n+1}^{4}\left(n^{3}+n^{2}\right)\).
参考答案
- 【答案】\(C\)
【解析】∵n是正奇数,
则\(7^{n}+C_{n}^{1} 7^{n-1}+C_{n}^{2} 7^{n-2}+\cdots \ldots+C_{n}^{n-1} 7\)
\(\begin{aligned} &=7^{n}+C_{n}^{1} 7^{n-1}+C_{n}^{2} 7^{n-2}+\cdots \ldots+C_{n}^{n-1} 7+C_{n}^{n}-1 \\ &=(7+1)^{n}-1 \\ &=(9-1)^{n}-1 \\ &=9^{n}-C_{n}^{1} 9^{n-1}+C_{n}^{2} 9^{n-2}-\ldots+C_{n}^{n-1} 9-C_{n}^{n}-1 \end{aligned}\),
\(∴\)它被\(9\)除的余数为\(-C_n^n-1=-2\),即它被\(9\)除的余数为\(7\),
故选:\(C\). - 【证明】\(11^{10}-1=(10+1)^{10}-1\)
\(\begin{aligned} &=\left(10^{10}+C_{10}^{1} \cdot 10^{9}+\cdots+C_{10}^{9} \cdot 10+1\right)-1 \\ &=10^{10}+C_{10}^{1} \cdot 10^{9}+C_{10}^{2} \cdot 10^{8}+\ldots+10^{2} \\ &=100\left(10^{8}+C_{10}^{1} \cdot 10^{7}+C_{10}^{2} \cdot 10^{6}+\ldots+1\right) \end{aligned}\).
\(\therefore 11^{10-1}\)能被\(100\)整除. - 【答案】\(1.17\)
【解析】\(1.02^{8}=(1+0.02)^{8}=1+C_{8}^{1} \times 0.02+C_{8}^{2} \times 0.02^{2}+\ldots\)\(\approx 1+0.16+0.0112 \approx 1.17\). - 【答案】\((3 n+2) \times 2^{n-1}\)
【解析】\(\because a_{n}=3 n+1\)为等差数列,\(\therefore a_{0}+a_{n}=a_{1}+a_{n-1}=\ldots\),
而\(C_{n}^{k}=C_{n}^{n-k}\),(运用反序求和方法),
\(\because W=C_{n}^{0}+4 C_{n}^{1}+7 C_{n}^{2}+\cdots+(3 n-2) C_{n}^{n-1}+(3 n+1) C_{n}^{n}\)①,
\(=(3 n+1) C_{n}^{n}+(3 n-2) C_{n}^{n-1}+(3 n-5) C_{n}^{n-2}+\cdots+4 C_{n}^{1}+C_{n}^{0}\)
\(\therefore W=(3 n+1) C_{n}^{0}+(3 n-2) C_{n}^{1}+(3 n-5) C_{n}^{n-2}+\cdots+4 C_{n}^{1}+C_{n}^{0}\)②,
①+②得\(2 W=(3 n+2)\left(C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+\cdots+C_{n}^{n}\right)=(3 n+2) \times 2^{n}\),
\(\therefore W=(3 n+2) \times 2^{n-1}\). - 【证明】由题意可得,\(k+1\)为正整数,即\(k\)为自然数,
\(\because\left(1+\dfrac{1}{k+1}\right)^{k+1}=C_{k+1}^{0}+C_{k+1}^{1} \cdot \dfrac{1}{k+1}+C_{k+1}^{2} \cdot\left(\dfrac{1}{k+1}\right)^{2}+\cdots+C_{k+1}^{k+1} \cdot\left(\dfrac{1}{k+1}\right)^{k+1}\)
\(\geq C_{k+1}^{0}+C_{k+1}^{1} \cdot \dfrac{1}{k+1}=1+1=2\),
当\(k=0\)时,取等号,
即\(\left(1+\dfrac{1}{k+1}\right)^{k+1} \geq 2\)成立. - 【答案】\(\text { (1) } f(1)=C_{n}^{3}\),\(f(2)=8C_n^3\),\(f(3)=27C_n^3\) \((2)\)见解析
【解析】(1)\(\because(a x+1)^{n}\)二项展开式中的\(x^{3}\)项的系数为\(C_n^3 a^3\).
\(∴f(a)=C_n^3 a^3\),则\(f(1)=C_n^3\),\(f(2)=8C_n^3\),\(f(3)=27C_n^3\);
证明:(2)由(1)得,\(\sum_{a=1}^{n} f(a)=C_{n}^{3}\left(1^{3}+2^{3}+\ldots+n^{3}\right)\).
首先利用数学归纳法证明
\(1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\dfrac{n^{2}(n+1)^{2}}{4}(n \geq 3)\).
①当\(n=3\)时,\(1^{3}+2^{3}+3^{3}=36=\dfrac{3^{2} \times 4^{2}}{4}\),
②假设当\(n=k(k≥3\)且\(k∈N^*)\)时,结论成立,
即\(1^{3}+2^{3}+\cdots+k^{3}=\dfrac{k^{2}(k+1)^{2}}{4}\).
那么,当\(n=k+1\)时,\(1^{3}+2^{3}+\ldots+k^{3}+(k+1)^{3}\)
\(=\dfrac{k^{2}(k+1)^{2}}{4}+(k+1)^{3}=(k+1)^{2} \times \dfrac{k^{2}+4 k+4}{4}=\dfrac{(k+1)^{2}(k+2)^{2}}{4}\)
\(∴\)对任意不小于\(3\)的正整数\(n\),均有\(1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\dfrac{n^{2}(n+1)^{2}}{4}\),
\(\therefore \sum_{a=1}^{n} f(a)=C_{n}^{3} \times \dfrac{n^{2}(n+1)^{2}}{4}=\dfrac{n(n-1)(n-2)}{6} \times \dfrac{n^{2}(n+1)^{2}}{4}\)
\(=\dfrac{(n-2)(n-1) n(n+1)}{24} \times\left(n^{3}+n^{2}\right)=C_{n+1}^{4}\left(n^{3}+n^{2}\right)\).
故\(\sum_{a=1}^{n} f(a)=C_{n+1}^{4}\left(n^{3}+n^{2}\right)\).