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IAL 2021 Oct Q9

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 9

题目

Problem

(a) Show that

n5−(n−1)5≡5n4−10n3+10n2−5n+1\begin{align*} n^5 - (n - 1)^5 \equiv 5n^4 - 10n^3 + 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 = \frac{1}{30} n(n + 1)(2n + 1)(an^2 + bn + c) \end{align*}

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

(7)

解答

(a)

解法一

思路

展开

这是一个恒等式证明。直接展开 (n−1)5(n-1)^5，再从 n5n^5 中减去它即可。因为题目是 “show that”，展开过程要写清楚，不能只写最后结果。

答题过程

展开

Using the binomial expansion,

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

This proves the identity.

(b)

解法一

思路

展开

由 (a) 可得

r5−(r−1)5=5r4−10r3+10r2−5r+1.\begin{align*} r^5-(r-1)^5 =5r^4-10r^3+10r^2-5r+1. \end{align*}

把 r=1r=1 到 r=nr=n 全部相加，左边会望远镜式抵消，只剩 n5n^5。然后使用已知求和公式

∑r,∑r2,∑r3\begin{align*} \sum r,\qquad \sum r^2,\qquad \sum r^3 \end{align*}

解出 ∑r4\sum r^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 =&\,5r^4-10r^3+10r^2-5r+1. \end{align*}

Sum this from r=1r=1 to r=nr=n:

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

The left hand side is a telescoping sum:

 (15−05) +(25−15) +(35−25) +⋯ +(n5−(n−1)5)= n5.\begin{align*} &\,\left(1^5-0^5\right)\\[4mm] &\,\hspace{2pt}+\left(2^5-1^5\right)\\[4mm] &\,\hspace{4pt}+\left(3^5-2^5\right)\\[4mm] &\,\hspace{6pt}+\cdots\\[4mm] &\,\hspace{8pt}+\left(n^5-(n-1)^5\right)\\[4mm] =&\,n^5. \end{align*}

So

n5= 5∑r=1nr4−10∑r=1nr3+10∑r=1nr2 −5∑r=1nr+∑r=1n1.\begin{align*} n^5 =&\,5\sum_{r=1}^{n}r^4 -10\sum_{r=1}^{n}r^3 +10\sum_{r=1}^{n}r^2\\[2mm] &\,\hspace{2pt}-5\sum_{r=1}^{n}r +\sum_{r=1}^{n}1. \end{align*}

Since

∑r=1n1=n,\begin{align*} \sum_{r=1}^{n}1=n, \end{align*}

we have

n5= 5∑r=1nr4−10∑r=1nr3+10∑r=1nr2 −5∑r=1nr+n.\begin{align*} n^5 =&\,5\sum_{r=1}^{n}r^4 -10\sum_{r=1}^{n}r^3 +10\sum_{r=1}^{n}r^2\\[2mm] &\,\hspace{2pt}-5\sum_{r=1}^{n}r+n. \end{align*}

Rearrange to make 5∑r45\sum r^4 the subject:

5∑r=1nr4= n5+10∑r=1nr3−10∑r=1nr2 +5∑r=1nr−n.\begin{align*} 5\sum_{r=1}^{n}r^4 =&\,n^5 +10\sum_{r=1}^{n}r^3 -10\sum_{r=1}^{n}r^2\\[2mm] &\,\hspace{2pt}+5\sum_{r=1}^{n}r-n. \end{align*}

Use the standard summation formulae:

∑r=1nr= 12n(n+1),∑r=1nr2= 16n(n+1)(2n+1),∑r=1nr3= 14n2(n+1)2.\begin{align*} \sum_{r=1}^{n}r =&\,\frac12n(n+1),\\[2mm] \sum_{r=1}^{n}r^2 =&\,\frac16n(n+1)(2n+1),\\[2mm] \sum_{r=1}^{n}r^3 =&\,\frac14n^2(n+1)^2. \end{align*}

Then

5∑r=1nr4= n5+10[14n2(n+1)2] −10[16n(n+1)(2n+1)] +5[12n(n+1)]−n.\begin{align*} 5\sum_{r=1}^{n}r^4 =&\,n^5 +10\left[\frac14n^2(n+1)^2\right]\\[2mm] &\,\hspace{2pt}-10\left[\frac16n(n+1)(2n+1)\right]\\[2mm] &\,\hspace{4pt}+5\left[\frac12n(n+1)\right]-n. \end{align*}

Simplify the constants:

5∑r=1nr4= n5+52n2(n+1)2 −53n(n+1)(2n+1) +52n(n+1)−n.\begin{align*} 5\sum_{r=1}^{n}r^4 =&\,n^5+\frac52n^2(n+1)^2\\[2mm] &\,\hspace{2pt}-\frac53n(n+1)(2n+1)\\[2mm] &\,\hspace{4pt}+\frac52n(n+1)-n. \end{align*}

Now expand enough to factor:

5∑r=1nr4= n5+52(n4+2n3+n2) −53(2n3+3n2+n) +52(n2+n)−n.\begin{align*} 5\sum_{r=1}^{n}r^4 =&\,n^5+\frac52(n^4+2n^3+n^2)\\[2mm] &\,\hspace{2pt}-\frac53(2n^3+3n^2+n)\\[2mm] &\,\hspace{4pt}+\frac52(n^2+n)-n. \end{align*}

Collect like terms:

5∑r=1nr4= n5+52n4+53n3 −16n.\begin{align*} 5\sum_{r=1}^{n}r^4 =&\,n^5+\frac52n^4+\frac53n^3\\[2mm] &\,\hspace{2pt}-\frac16n. \end{align*}

Therefore

∑r=1nr4= 15n5+12n4+13n3−130n.\begin{align*} \sum_{r=1}^{n}r^4 =&\,\frac15n^5+\frac12n^4+\frac13n^3-\frac1{30}n. \end{align*}

Put over a common denominator:

∑r=1nr4= 6n5+15n4+10n3−n30= n(6n4+15n3+10n2−1)30.\begin{align*} \sum_{r=1}^{n}r^4 =&\,\frac{6n^5+15n^4+10n^3-n}{30}\\[2mm] =&\,\frac{n(6n^4+15n^3+10n^2-1)}{30}. \end{align*}

Now factor the quartic:

6n4+15n3+10n2−1= (n+1)(2n+1)(3n2+3n−1).\begin{align*} 6n^4+15n^3+10n^2-1 =&\,(n+1)(2n+1)(3n^2+3n-1). \end{align*}

Hence

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

Therefore

a=3,b=3,c=−1.\begin{align*} a=3,\qquad b=3,\qquad c=-1. \end{align*}