Question

Problem

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

$\sum_{r=1}^{n} r^2(2r-1) = \frac{1}{6}n(n+1)(3n^2+n-1)$

(5)

中文翻译

用数学归纳法证明，对于 $n \in \mathbb{Z}^+$

$\sum_{r=1}^{n} r^2(2r-1) = \frac{1}{6}n(n+1)(3n^2+n-1)$

(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

\frac16(1)(1+1)\big(3(1)^2+1-1\big)=1.

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

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

S_k=\frac16k(k+1)(3k^2+k-1).

Therefore,

\begin{align*} S_{k+1} =&\,S_k+(k+1)^2\big(2(k+1)-1\big)\\ =&\,\frac16k(k+1)(3k^2+k-1)\\ &\,+(k+1)^2(2k+1)\\ =&\,\frac16(k+1) \big(3k^3+13k^2+17k+6\big)\\ =&\,\frac16(k+1)(k+2)(3k^2+7k+3). \end{align*}

Since

3k^2+7k+3 =3(k+1)^2+(k+1)-1,

we obtain

\begin{align*} S_{k+1} =&\,\frac16(k+1)\big((k+1)+1\big)\\ &\,\times\big(3(k+1)^2+(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{aligned} \sum_{r=1}^{n}r^2(2r-1) =&\,\frac16n(n+1)\\ &\,\times(3n^2+n-1) \end{aligned} }

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