题目

Problem

Prove by induction that, for $n \in \mathbb{Z}^+$

\begin{align*} &\,\sum_{r=1}^{n}r^2(2r-1)\\ =&\,\frac16n(n+1)\\ &\,\hspace{2pt}\times(3n^2+n-1) \end{align*}

(5)

题目中文翻译

用数学归纳法证明，对于 $n \in \mathbb{Z}^+$

\begin{align*} &\,\sum_{r=1}^{n}r^2(2r-1)\\ =&\,\frac16n(n+1)\\ &\,\hspace{2pt}\times(3n^2+n-1) \end{align*}

(5)

解答

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先验证 $n=1$ 。然后假设结论在 $n=k$ 时成立，把第 $k+1$ 项 $(k+1)^2(2k+1)$ 加到前 $k$ 项的和上，并把结果因式分解成命题在 $n=k+1$ 时应有的形式。

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Let

S_n=\sum_{r=1}^{n}r^2(2r-1).

When $n=1$ ,

S_1=1^2(2-1)=1,

while

\begin{align*} &\,\frac16(1)(1+1)\\ &\,\hspace{2pt}\times\big(3(1)^2+1-1\big)\\ =&\,1. \end{align*}

Thus the result is true for $n=1$ .

Assume that the result is true for $n=k$ , where $k\in\mathbb Z^+$ . Then

\begin{align*} S_k=&\,\frac16k(k+1)\\ &\,\hspace{2pt}\times(3k^2+k-1). \end{align*}

Therefore,

\begin{align*} &\,S_{k+1}\\ =&\,S_k+(k+1)^2\\ &\,\hspace{2pt}\times\big(2(k+1)-1\big)\\ =&\,\frac16k(k+1)(3k^2+k-1)\\ &\,\hspace{2pt}+(k+1)^2(2k+1)\\ =&\,\frac16(k+1)\Big[k(3k^2+k-1)\\ &\,\hspace{2pt}+6(k+1)(2k+1)\Big]. \end{align*}

The expression in square brackets simplifies to

\begin{align*} &\,k(3k^2+k-1)\\ &\,\hspace{2pt}+6(k+1)(2k+1)\\ =&\,3k^3+k^2-k\\ &\,\hspace{2pt}+12k^2+18k+6\\ =&\,3k^3+13k^2+17k+6\\ =&\,3k^3+6k^2+7k^2\\ &\,\hspace{2pt}+14k+3k+6\\ =&\,3k^2(k+2)+7k(k+2)\\ &\,\hspace{2pt}+3(k+2)\\ =&\,(k+2)(3k^2+7k+3). \end{align*}

Hence,

\begin{align*} S_{k+1}=&\,\frac16(k+1)(k+2)\\ &\,\hspace{2pt}\times(3k^2+7k+3). \end{align*}

Also,

\begin{align*} &\,3(k+1)^2\\ &\,\hspace{2pt}+(k+1)-1\\ =&\,3(k^2+2k+1)\\ &\,\hspace{2pt}+k+1-1\\ =&\,3k^2+6k+3+k\\ =&\,3k^2+7k+3. \end{align*}

Therefore,

\begin{align*} &\,S_{k+1}\\ =&\,\frac16(k+1)\big((k+1)+1\big)\\ &\,\hspace{2pt}\times\Big[3(k+1)^2\\ &\,\hspace{4pt}+(k+1)-1\Big]. \end{align*}

Hence the result is true for $n=k+1$ whenever it is true for $n=k$ . Since it is true for $n=1$ , by mathematical induction,

\boxed{ \begin{align*} &\,\sum_{r=1}^{n}r^2(2r-1)\\ =&\,\frac16n(n+1)\\ &\,\hspace{2pt}\times(3n^2+n-1) \end{align*} }

for all $n\in\mathbb Z^+$ .