题目

查看数据源：599165b92bfec200011de9ca

设 $d$ 为非零实数，${a_n} = \dfrac{1}{n}\left[ {{\mathrm{C}}_n^1d + 2{\mathrm{C}}_n^2{d^2} + \cdots + \left(n - 1\right){\mathrm{C}}_n^{n - 1}{d^{n - 1}} + n{\mathrm{C}}_n^n{d^n}} \right]\left( {n \in {{\mathbb{N}}^ * }} \right)$．

【难度】

【出处】

2011年高考四川卷（理）

【标注】

写出 ${a_1}$，${a_2}$，${a_3}$ 并判断 $\left\{ {a_n}\right\} $ 是否为等比数列．若是，给出证明；若不是，说明理由；
标注
答案

解析

由已知可得\[{a_1} = d,{a_2} = d\left( {1 + d} \right),{a_3} = d{\left( {1 + d} \right)^2}.\]当 $n \geqslant 2,k \geqslant 1$ 时，$\dfrac{k}{n}{\mathrm{C}}_n^k = {\mathrm{C}}_{n - 1}^{k - 1}$，因此\[\begin{split}{a_n} & = \sum\limits_{k = 1}^n {\dfrac{k}{n}} {\mathrm{C}}_n^k{d^k} = \sum\limits_{k = 1}^n {{\mathrm{C}}_{n - 1}^{k - 1}{d^k}} \\& = d\sum\limits_{k = 0}^{n - 1} {{\mathrm{C}}_{n - 1}^k{d^k}} \\& = d{\left( {d + 1} \right)^{n - 1}}.\end{split}\]由此可见，
当 $d \ne - 1$ 时，$\left\{ {a_n} \right\}$ 是以 $d$ 为首项，$d + 1$ 为公比的等比数列；
当 $d = - 1$ 时，${a_1} = - 1$，${a_n} = 0\left( {n \geqslant 2} \right)$，此时 $\left\{ {a_n} \right\}$ 不是等比数列．
设 ${b_n} = nd{a_n}\left(n \in {{\mathbb{N}}^ * }\right)$，求数列 $\left\{ {b_n}\right\} $ 的前 $n$ 项和 ${S_n}$．
标注
答案

解析

由（1）可知，${a_n} = d{\left( {d + 1} \right)^{n - 1}}$，从而 ${b_n} = n{d^2}{\left( {d + 1} \right)^{n - 1}}$，\[{S_n} = {d^2}\left[ {1 + 2\left( {d + 1} \right) + 3{{\left( {d + 1} \right)}^2} + \cdots + \left( {n - 1} \right){{\left( {d + 1} \right)}^{n - 2}} + n{{\left( {d + 1} \right)}^{n - 1}}} \right], \quad \cdots \cdots ① \]当 $d = - 1$ 时，\[{S_n} = {d^2} = 1;\]当 $d \ne - 1$ 时，① 式两边同乘 $\left(d + 1\right)$ 得\[\left( {d + 1} \right){S_n} = {d^2}\left[ {\left( {d + 1} \right) + 2{{\left( {d + 1} \right)}^2} + 3{{\left( {d + 1} \right)}^3} + \cdots + \left( {n - 1} \right){{\left( {d + 1} \right)}^{n - 1}} + n{{\left( {d + 1} \right)}^n}} \right], \quad \cdots \cdots ② \]①，② 式相减可得\[ \begin{split}- d{S_n} & = {d^2}\left[ {1 + \left( {d + 1} \right) + {{\left( {d + 1} \right)}^2} + \cdots + {{\left( {d + 1} \right)}^{n - 1}} - n{{\left( {d + 1} \right)}^n}} \right] \\& = {d^2}\left[ {\dfrac{{{{\left( {d + 1} \right)}^n} - 1}}{d} - n{{\left( {d + 1} \right)}^n}} \right],\end{split}\]化简即得 ${S_n} = {\left( {d + 1} \right)^n}\left( {nd - 1} \right) + 1$．
综上可知，\[{S_n} = {\left( {d + 1} \right)^n}\left( {nd - 1} \right) + 1.\]

题目问题1 答案1 解析1 备注1 问题2 答案2 解析2

由（1）可知，${a_n} = d{\left( {d + 1} \right)^{n - 1}}$，从而 ${b_n} = n{d^2}{\left( {d + 1} \right)^{n - 1}}$，\[{S_n} = {d^2}\left[ {1 + 2\left( {d + 1} \right) + 3{{\left( {d + 1} \right)}^2} + \cdots + \left( {n - 1} \right){{\left( {d + 1} \right)}^{n - 2}} + n{{\left( {d + 1} \right)}^{n - 1}}} \right], \quad \cdots \cdots ① \]当 $d = - 1$ 时，\[{S_n} = {d^2} = 1;\]当 $d \ne - 1$ 时，① 式两边同乘 $\left(d + 1\right)$ 得\[\left( {d + 1} \right){S_n} = {d^2}\left[ {\left( {d + 1} \right) + 2{{\left( {d + 1} \right)}^2} + 3{{\left( {d + 1} \right)}^3} + \cdots + \left( {n - 1} \right){{\left( {d + 1} \right)}^{n - 1}} + n{{\left( {d + 1} \right)}^n}} \right], \quad \cdots \cdots ② \]①，② 式相减可得\[ \begin{split}- d{S_n} & = {d^2}\left[ {1 + \left( {d + 1} \right) + {{\left( {d + 1} \right)}^2} + \cdots + {{\left( {d + 1} \right)}^{n - 1}} - n{{\left( {d + 1} \right)}^n}} \right] \\& = {d^2}\left[ {\dfrac{{{{\left( {d + 1} \right)}^n} - 1}}{d} - n{{\left( {d + 1} \right)}^n}} \right],\end{split}\]化简即得 ${S_n} = {\left( {d + 1} \right)^n}\left( {nd - 1} \right) + 1$．<br/>综上可知，\[{S_n} = {\left( {d + 1} \right)^n}\left( {nd - 1} \right) + 1.\]