CalcGospel 國際數學圖譜
返回

FP1 2022 Jan Q9

A Level / Edexcel / FP1

IAL 2022 Jan Paper · Question 9

题目

Problem

(a) Prove by induction that, for n∈Nn \in \mathbb N

 ∑r=1nr3= 14n2(n+1)2\begin{align*} &\,\sum_{r=1}^{n}r^3\\ =&\,\frac14n^2(n+1)^2 \end{align*}
(5)

(b) Using the standard summation formulae, show that

 ∑r=1nr(r+1)(r−1)= 14n(n+A) ×(n+B)(n+C)\begin{align*} &\,\sum_{r=1}^{n}r(r+1)(r-1)\\ =&\,\frac14n(n+A)\\ &\,\hspace{2pt}\times(n+B)(n+C) \end{align*}

where AA, BB and CC are constants to be determined.

(4)

(c) Determine the value of nn for which

 3∑r=1nr(r+1)(r−1)= 17∑r=n2nr2\begin{align*} &\,3\sum_{r=1}^{n}r(r+1)(r-1)\\ =&\,17\sum_{r=n}^{2n}r^2 \end{align*}
(5)
题目中文翻译

(a) 用数学归纳法证明：对于 n∈Nn \in \mathbb N，

 ∑r=1nr3= 14n2(n+1)2\begin{align*} &\,\sum_{r=1}^{n}r^3\\ =&\,\frac14n^2(n+1)^2 \end{align*}
(5)

(b) 使用标准求和公式证明

 ∑r=1nr(r+1)(r−1)= 14n(n+A) ×(n+B)(n+C)\begin{align*} &\,\sum_{r=1}^{n}r(r+1)(r-1)\\ =&\,\frac14n(n+A)\\ &\,\hspace{2pt}\times(n+B)(n+C) \end{align*}

其中 AA、BB、CC 为待确定的常数。

(4)

(c) 确定使下式成立的 nn 值

 3∑r=1nr(r+1)(r−1)= 17∑r=n2nr2\begin{align*} &\,3\sum_{r=1}^{n}r(r+1)(r-1)\\ =&\,17\sum_{r=n}^{2n}r^2 \end{align*}
(5)

解答

(a)

解法一

思路

展开

先验证 n=1n=1。归纳步骤中，在前 kk 项立方和后加上 (k+1)3(k+1)^3，再提取 14(k+1)2\frac14(k+1)^2；括号内恰好是 (k+2)2(k+2)^2，从而得到 n=k+1n=k+1 时的目标形式。

答题过程

展开

For n=1n=1,

∑r=11r3=1\sum_{r=1}^{1}r^3=1

and

14(1)2(1+1)2= 14×1×4= 1.\begin{align*} \frac14(1)^2(1+1)^2 =&\,\frac14\times1\times4\\ =&\,1. \end{align*}

Hence the result is true for n=1n=1.

Assume that the result is true for n=kn=k, where k∈Z+k\in\mathbb Z^+. Thus

 ∑r=1kr3= 14k2(k+1)2.\begin{align*} &\,\sum_{r=1}^{k}r^3\\ =&\,\frac14k^2(k+1)^2. \end{align*}

Then

 ∑r=1k+1r3= ∑r=1kr3+(k+1)3= 14k2(k+1)2 +(k+1)3.\begin{align*} &\,\sum_{r=1}^{k+1}r^3\\ =&\,\sum_{r=1}^{k}r^3+(k+1)^3\\ =&\,\frac14k^2(k+1)^2\\ &\,\hspace{2pt}+(k+1)^3. \end{align*}

Taking out the common factor 14(k+1)2\frac14(k+1)^2, write

 ∑r=1k+1r3= 14(k+1)2H,\begin{align*} &\,\sum_{r=1}^{k+1}r^3\\ =&\,\frac14(k+1)^2H, \end{align*}

where

H= k2+4(k+1)= k2+4k+4= k2+2k+2k+4= k(k+2)+2(k+2)= (k+2)(k+2)= (k+2)2.\begin{align*} H =&\,k^2+4(k+1)\\ =&\,k^2+4k+4\\ =&\,k^2+2k+2k+4\\ =&\,k(k+2)+2(k+2)\\ =&\,(k+2)(k+2)\\ =&\,(k+2)^2. \end{align*}

Therefore,

 ∑r=1k+1r3= 14(k+1)2(k+2)2= 14(k+1)2 ×((k+1)+1)2.\begin{align*} &\,\sum_{r=1}^{k+1}r^3\\ =&\,\frac14(k+1)^2(k+2)^2\\ =&\,\frac14(k+1)^2\\ &\,\hspace{2pt}\times\big((k+1)+1\big)^2. \end{align*}

This is the required result for n=k+1n=k+1. Since the result is true for n=1n=1, and truth for n=kn=k implies truth for n=k+1n=k+1, it follows by mathematical induction that

 ∑r=1nr3= 14n2(n+1)2\boxed{ \begin{aligned} &\,\sum_{r=1}^{n}r^3\\ =&\,\frac14n^2(n+1)^2 \end{aligned} }

for all n∈Z+n\in\mathbb Z^+.

(b)

解法一

思路

展开

先展开被加数：

 r(r+1)(r−1)= r3−r.\begin{align*} &\,r(r+1)(r-1)\\ =&\,r^3-r. \end{align*}

再使用 (a) 的立方和结果与标准的一次方求和公式。提取共同因子后，把剩余二次式分解为 (n−1)(n+2)(n-1)(n+2)。

答题过程

展开

First,

 r(r+1)(r−1)= r(r2−1)= r3−r.\begin{align*} &\,r(r+1)(r-1)\\ =&\,r(r^2-1)\\ =&\,r^3-r. \end{align*}

Write

S=∑r=1nr(r+1)(r−1).S=\sum_{r=1}^{n}r(r+1)(r-1).

Using part (a) and the standard result for the sum of the first nn positive integers,

S= ∑r=1nr3 −∑r=1nr= 14n2(n+1)2 −12n(n+1).\begin{align*} S =&\,\sum_{r=1}^{n}r^3\\ &\,\hspace{2pt}-\sum_{r=1}^{n}r\\ =&\,\frac14n^2(n+1)^2\\ &\,\hspace{2pt}-\frac12n(n+1). \end{align*}

Taking out the common factor gives

S=14n(n+1)K,S=\frac14n(n+1)K,

where

K= n(n+1)−2= n2+n−2= n2+2n−n−2= n(n+2)−(n+2)= (n−1)(n+2).\begin{align*} K =&\,n(n+1)-2\\ =&\,n^2+n-2\\ =&\,n^2+2n-n-2\\ =&\,n(n+2)-(n+2)\\ =&\,(n-1)(n+2). \end{align*}

Hence

S= 14n(n+1) ×(n−1)(n+2).\begin{align*} S =&\,\frac14n(n+1)\\ &\,\hspace{2pt}\times(n-1)(n+2). \end{align*}

Therefore one valid assignment of the constants is

A=1,B=−1,C=2,\begin{align*} \boxed{A=1},\\ \boxed{B=-1},\\ \boxed{C=2}, \end{align*}

where the order of A,B,CA,B,C may be interchanged.

(c)

解法一

思路

展开

先把从 nn 到 2n2n 的平方和写成“前 2n2n 项平方和减去前 n−1n-1 项平方和”，并用标准公式化简。然后代入 (b) 的结果，约去正整数条件下非零的公共因子，解所得二次方程并排除非正整数根。

答题过程

展开

Write

T=∑r=n2nr2.T=\sum_{r=n}^{2n}r^2.

Using the standard result for the sum of squares,

T= ∑r=12nr2 −∑r=1n−1r2= (2n)(2n+1)(4n+1)6 −(n−1)n(2n−1)6.\begin{align*} T =&\,\sum_{r=1}^{2n}r^2\\ &\,\hspace{2pt}-\sum_{r=1}^{n-1}r^2\\ =&\,\frac{(2n)(2n+1)(4n+1)}6\\ &\,\hspace{2pt}-\frac{(n-1)n(2n-1)}6. \end{align*}

Taking out the common factor n/6n/6, we have

T=n6L,T=\frac n6L,

where

L= 2(2n+1)(4n+1) −(n−1)(2n−1)= 2(8n2+6n+1) −(2n2−3n+1)= 16n2+12n+2 −2n2+3n−1= 14n2+15n+1= 14n2+14n+n+1= 14n(n+1)+(n+1)= (14n+1)(n+1).\begin{align*} L =&\,2(2n+1)(4n+1)\\ &\,\hspace{2pt}-(n-1)(2n-1)\\ =&\,2(8n^2+6n+1)\\ &\,\hspace{2pt}-(2n^2-3n+1)\\ =&\,16n^2+12n+2\\ &\,\hspace{2pt}-2n^2+3n-1\\ =&\,14n^2+15n+1\\ =&\,14n^2+14n+n+1\\ =&\,14n(n+1)+(n+1)\\ =&\,(14n+1)(n+1). \end{align*}

Thus

T=n6(14n+1)(n+1).T=\frac n6(14n+1)(n+1).

Using part (b), the given equation 3S=17T3S=17T becomes

 34n(n+1)(n−1) ×(n+2)= 176n(n+1) ×(14n+1).\begin{align*} &\,\frac34n(n+1)(n-1)\\ &\,\hspace{2pt}\times(n+2)\\ =&\,\frac{17}{6}n(n+1)\\ &\,\hspace{2pt}\times(14n+1). \end{align*}

Since nn is a positive integer, n(n+1)≠0n(n+1)\ne0, so this factor may be cancelled. Multiplying the remaining equation by 1212 gives

 9(n−1)(n+2)= 34(14n+1).\begin{align*} &\,9(n-1)(n+2)\\ =&\,34(14n+1). \end{align*}

Expanding both sides,

 9n2+9n−18= 476n+34.\begin{align*} &\,9n^2+9n-18\\ =&\,476n+34. \end{align*}

Rearranging and factorising,

 0= 9n2+9n−18 −476n−34= 9n2−467n−52= 9n2−468n+n−52= 9n(n−52)+(n−52)= (9n+1)(n−52).\begin{align*} &\,0\\ =&\,9n^2+9n-18\\ &\,\hspace{2pt}-476n-34\\ =&\,9n^2-467n-52\\ =&\,9n^2-468n+n-52\\ =&\,9n(n-52)+(n-52)\\ =&\,(9n+1)(n-52). \end{align*}

Thus n=−1/9n=-1/9 or n=52n=52. Since n∈Z+n\in\mathbb Z^+,

n=52.\boxed{n=52}.