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FP2 2021 Oct Q9

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 9

题目

Problem

9.

(a) Show that

 n5−(n−1)5≡ 5n4−10n3 +10n2−5n+1\begin{align*} &\,n^5-(n-1)^5\\[4mm] \equiv&\,5n^4-10n^3\\[4mm] &\,\hspace{2pt}+10n^2-5n+1 \end{align*}
(2)

(b) Hence, using the method of differences, show that for all integer values of nn,

 ∑r=1nr4= 130n(n+1)(2n+1) ×(an2+bn+c)\begin{align*} &\,\sum_{r=1}^{n}r^4\\[4mm] =&\,\frac{1}{30}n(n+1)(2n+1)\\[4mm] &\,\hspace{2pt}\times(an^2+bn+c) \end{align*}

where aa, bb and cc are integers to be determined.

(7)
题目中文翻译

9. (a) 证明

 n5−(n−1)5≡ 5n4−10n3 +10n2−5n+1.\begin{align*} &\,n^5-(n-1)^5\\[4mm] \equiv&\,5n^4-10n^3\\[4mm] &\,\hspace{2pt}+10n^2-5n+1. \end{align*}
(2)

(b) 由此，使用差分法证明，对所有整数 nn，

 ∑r=1nr4= 130n(n+1)(2n+1) ×(an2+bn+c),\begin{align*} &\,\sum_{r=1}^{n}r^4\\[4mm] =&\,\frac{1}{30}n(n+1)(2n+1)\\[4mm] &\,\hspace{2pt}\times(an^2+bn+c), \end{align*}

其中 aa、bb 和 cc 是待求的整数。

(7)

解答

(a)

解法一

思路

展开

先用二项式定理完整展开 (n−1)5(n-1)^5，再从 n5n^5 中减去它。注意括号前的负号会改变括号内每一项的符号。

答题过程

展开

Using the binomial expansion,

 (n−1)5= n5−5n4+10n3 −10n2+5n−1.\begin{align*} &\,(n-1)^5\\[4mm] =&\,n^5-5n^4+10n^3\\[4mm] &\,\hspace{2pt}-10n^2+5n-1. \end{align*}

Therefore,

 n5−(n−1)5= n5−[n5−5n4 +10n3−10n2 +5n−1]= n5−n5+5n4 −10n3+10n2 −5n+1= 5n4−10n3 +10n2−5n+1.\begin{align*} &\,n^5-(n-1)^5\\[4mm] =&\,n^5-\bigg[n^5-5n^4\\[4mm] &\,\hspace{2pt}+10n^3-10n^2\\[4mm] &\,\hspace{4pt}+5n-1\bigg]\\[4mm] =&\,n^5-n^5+5n^4\\[4mm] &\,\hspace{2pt}-10n^3+10n^2\\[4mm] &\,\hspace{4pt}-5n+1\\[4mm] =&\,5n^4-10n^3\\[4mm] &\,\hspace{2pt}+10n^2-5n+1. \end{align*}

This proves the identity.

(b)

解法一

思路

展开

把 (a) 中的 nn 换成 rr，再把 r=1r=1 到 r=nr=n 的等式相加。左边相邻的五次方项会抵消，只剩 n5n^5。右边则可利用一次、二次、三次幂的求和公式，解出四次幂的和。

为缩短记号，对固定的 nn，记

Sk= ∑r=1nrk.\begin{align*} S_k=&\,\sum_{r=1}^{n}r^k. \end{align*}

这里 S1S_1、S2S_2、S3S_3 的公式已知，要求的是 S4S_4。

答题过程

展开

From part (a),

 r5−(r−1)5= 5r4−10r3 +10r2−5r+1.\begin{align*} &\,r^5-(r-1)^5\\[4mm] =&\,5r^4-10r^3\\[4mm] &\,\hspace{2pt}+10r^2-5r+1. \end{align*}

For this fixed nn, write

Sk= ∑r=1nrk.\begin{align*} S_k=&\,\sum_{r=1}^{n}r^k. \end{align*}

Sum the identity from r=1r=1 to r=nr=n. The left-hand side is

 ∑r=1n[r5−(r−1)5]= (15−05) +(25−15) +(35−25) +⋯ +[(n−1)5−(n−2)5] +[n5−(n−1)5]= n5.\begin{align*} &\,\sum_{r=1}^{n}\big[r^5-(r-1)^5\big]\\[4mm] =&\,(1^5-0^5)\\[4mm] &\,\hspace{2pt}+(2^5-1^5)\\[4mm] &\,\hspace{4pt}+(3^5-2^5)\\[4mm] &\,\hspace{6pt}+\cdots\\[4mm] &\,\hspace{2pt}+\big[(n-1)^5-(n-2)^5\big]\\[4mm] &\,\hspace{4pt}+\big[n^5-(n-1)^5\big]\\[4mm] =&\,n^5. \end{align*}

On the right-hand side, the sum of the constant terms is nn. Hence

n5= 5S4−10S3 +10S2−5S1+n.\begin{align*} n^5=&\,5S_4-10S_3\\[4mm] &\,\hspace{2pt}+10S_2-5S_1+n. \end{align*}

Rearranging,

5S4= n5+10S3 −10S2+5S1−n.\begin{align*} 5S_4=&\,n^5+10S_3\\[4mm] &\,\hspace{2pt}-10S_2+5S_1-n. \end{align*}

The standard summation formulae give

S1= 12n(n+1),S2= 16n(n+1)(2n+1),S3= 14n2(n+1)2.\begin{align*} S_1=&\,\frac12n(n+1),\\[4mm] S_2=&\,\frac16n(n+1)(2n+1),\\[4mm] S_3=&\,\frac14n^2(n+1)^2. \end{align*}

Substituting,

 5S4= n5 +10[14n2(n+1)2] −10[16n(n+1) ×(2n+1)] +5[12n(n+1)]−n.\begin{align*} &\,5S_4\\[4mm] =&\,n^5\\[4mm] &\,\hspace{2pt}+10\bigg[\frac14n^2(n+1)^2\bigg]\\[4mm] &\,\hspace{4pt}-10\bigg[\frac16n(n+1)\\[4mm] &\,\hspace{6pt}\times(2n+1)\bigg]\\[4mm] &\,\hspace{6pt}+5\bigg[\frac12n(n+1)\bigg]-n. \end{align*}

Simplifying the numerical coefficients,

 5S4= n5+52n2(n+1)2 −53n(n+1)(2n+1) +52n(n+1)−n.\begin{align*} &\,5S_4\\[4mm] =&\,n^5+\frac52n^2(n+1)^2\\[4mm] &\,\hspace{2pt}-\frac53n(n+1)(2n+1)\\[4mm] &\,\hspace{4pt}+\frac52n(n+1)-n. \end{align*}

Expand:

 5S4= n5+52(n4+2n3+n2) −53(2n3+3n2+n) +52(n2+n)−n.\begin{align*} &\,5S_4\\[4mm] =&\,n^5+\frac52(n^4+2n^3+n^2)\\[4mm] &\,\hspace{2pt}-\frac53(2n^3+3n^2+n)\\[4mm] &\,\hspace{4pt}+\frac52(n^2+n)-n. \end{align*}

Collecting like terms,

5S4= n5+52n4 +53n3−16n.\begin{align*} 5S_4=&\,n^5+\frac52n^4\\[4mm] &\,\hspace{2pt}+\frac53n^3-\frac16n. \end{align*}

Therefore,

S4= 15n5+12n4 +13n3−130n.\begin{align*} S_4=&\,\frac15n^5+\frac12n^4\\[4mm] &\,\hspace{2pt}+\frac13n^3-\frac1{30}n. \end{align*}

To combine the terms, define the numerator NN by

N= 6n5+15n4 +10n3−n.\begin{align*} N=&\,6n^5+15n^4\\[4mm] &\,\hspace{2pt}+10n^3-n. \end{align*}

Then S4=N/30S_4=N/30. First take out a factor of nn:

N= nF,F= 6n4+15n3 +10n2−1.\begin{align*} N=&\,nF,\\[4mm] F=&\,6n^4+15n^3\\[4mm] &\,\hspace{2pt}+10n^2-1. \end{align*}

To factor FF, group its terms:

F= (6n4+6n3) +(9n3+9n2) +(n2−1)= 6n3(n+1) +9n2(n+1) +(n−1)(n+1).\begin{align*} F=&\,(6n^4+6n^3)\\[4mm] &\,\hspace{2pt}+(9n^3+9n^2)\\[4mm] &\,\hspace{4pt}+(n^2-1)\\[4mm] =&\,6n^3(n+1)\\[4mm] &\,\hspace{2pt}+9n^2(n+1)\\[4mm] &\,\hspace{4pt}+(n-1)(n+1). \end{align*}

Thus F=(n+1)GF=(n+1)G, where

G= 6n3+9n2+n−1.\begin{align*} G=&\,6n^3+9n^2+n-1. \end{align*}

Now group the terms of GG to take out 2n+12n+1:

G= (6n3+3n2) +(6n2+3n) −(2n+1)= 3n2(2n+1) +3n(2n+1) −(2n+1)= (2n+1) ×(3n2+3n−1).\begin{align*} G=&\,(6n^3+3n^2)\\[4mm] &\,\hspace{2pt}+(6n^2+3n)\\[4mm] &\,\hspace{4pt}-(2n+1)\\[4mm] =&\,3n^2(2n+1)\\[4mm] &\,\hspace{2pt}+3n(2n+1)\\[4mm] &\,\hspace{4pt}-(2n+1)\\[4mm] =&\,(2n+1)\\[4mm] &\,\hspace{2pt}\times(3n^2+3n-1). \end{align*}

Hence

S4= 130n(n+1)(2n+1) ×(3n2+3n−1).\begin{align*} S_4=&\,\frac1{30}n(n+1)(2n+1)\\[4mm] &\,\hspace{2pt}\times(3n^2+3n-1). \end{align*}

In the notation of the question,

 ∑r=1nr4= 130n(n+1)(2n+1) ×(3n2+3n−1).\begin{align*} &\,\sum_{r=1}^{n}r^4\\[4mm] =&\,\frac1{30}n(n+1)(2n+1)\\[4mm] &\,\hspace{2pt}\times(3n^2+3n-1). \end{align*}

Comparing coefficients gives

a= 3,b= 3,c= −1.\begin{align*} a=&\,3,\\[4mm] b=&\,3,\\[4mm] c=&\,-1. \end{align*}