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FP1 2020 Oct Q4

A Level / Edexcel / FP1

IAL 2020 Oct Paper · Question 4

题目

Problem

4. (a) Use the standard results for ∑r=1nr2\displaystyle\sum_{r=1}^{n} r^2 and ∑r=1nr\displaystyle\sum_{r=1}^{n} r to show that

 ∑r=1n(2r−1)2= 13n(4n2−1)\begin{align*} &\,\sum_{r=1}^{n}(2r-1)^2 \\ =&\,\frac13n(4n^2-1) \end{align*}

for all positive integers nn.

(5)

(b) Hence find the exact value of the sum of the squares of the odd numbers between 200200 and 500500

(4)
题目中文翻译

4. (a) 利用 ∑r=1nr2\displaystyle\sum_{r=1}^{n} r^2 和 ∑r=1nr\displaystyle\sum_{r=1}^{n} r 的标准结果证明

 ∑r=1n(2r−1)2= 13n(4n2−1)\begin{align*} &\,\sum_{r=1}^{n}(2r-1)^2 \\ =&\,\frac13n(4n^2-1) \end{align*}

对所有正整数 nn 成立。

(5)

(b) 由此求 200200 到 500500 之间所有奇数的平方和的精确值。

(4)

解答

(a)

解法一

思路

展开

先展开每项，再分别使用平方和、自然数和及常数项求和。提取公因式后，把括号内的乘积逐项展开并合并，得到题目要求的结果。

答题过程

展开

Let

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

Since

(2r−1)2=4r2−4r+1,(2r-1)^2=4r^2-4r+1,

we have

Sn= 4∑r=1nr2− 4∑r=1nr+n.\begin{align*} S_n=&\,4\sum_{r=1}^{n}r^2 \\ -&\,4\sum_{r=1}^{n}r+n. \end{align*}

Using the standard results,

Sn= 23n(n+1)(2n+1)− 2n(n+1)+n.\begin{align*} S_n=&\,\frac{2}{3}n(n+1)(2n+1) \\ -&\,2n(n+1)+n. \end{align*}

Factoring out 13n\frac13n gives

Sn= 13n[2(n+1)(2n+1) −6(n+1)+3]= 13n[2(2n2+3n+1) −6n−6+3]= 13n[4n2+6n+2 −6n−6+3]= 13n(4n2−1).\begin{align*} S_n=&\,\frac13n\big[2(n+1)(2n+1) \\ &\,\hspace{2pt}-6(n+1)+3\big] \\ =&\,\frac13n\big[2(2n^2+3n+1) \\ &\,\hspace{2pt}-6n-6+3\big] \\ =&\,\frac13n\big[4n^2+6n+2 \\ &\,\hspace{2pt}-6n-6+3\big] \\ =&\,\frac13n(4n^2-1). \end{align*}

Hence, for every positive integer nn,

 ∑r=1n(2r−1)2= 13n(4n2−1).\begin{align*} &\,\sum_{r=1}^{n}(2r-1)^2 \\ =&\,\boxed{\frac13n(4n^2-1)}. \end{align*}

(b)

解法一

思路

展开

把奇数写成 2r−12r-1，则最小的 201201 对应 r=101r=101，最大的 499499 对应 r=250r=250。利用 (a) 的结果，以前 250250 项和减去前 100100 项和。

答题过程

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The first and last odd numbers in the interval are 201201 and 499499. Solving

2r−1=2012r-1=201

gives r=101r=101, while

2r−1=4992r-1=499

gives r=250r=250. Therefore, using SnS_n from part (a), the required sum is

 ∑r=101250(2r−1)2= S250−S100.\begin{align*} &\,\sum_{r=101}^{250}(2r-1)^2 \\ =&\,S_{250}-S_{100}. \end{align*}

From part (a),

S250= 13(250)[4(250)2−1]= 20 833 250,\begin{align*} S_{250} =&\,\frac13(250)\big[4(250)^2-1\big] \\ =&\,20\,833\,250, \end{align*}

and

S100= 13(100)[4(100)2−1]= 1 333 300.\begin{align*} S_{100} =&\,\frac13(100)\big[4(100)^2-1\big] \\ =&\,1\,333\,300. \end{align*}

Hence the required sum is

 20 833 250−1 333 300= 19 499 950.\begin{align*} &\,20\,833\,250-1\,333\,300 \\ =&\,\boxed{19\,499\,950}. \end{align*}