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IAL 2026 Jan FP1 Q8

A Level / Edexcel / FP1

IAL 2026 Jan Paper · Question 8

Question

Problem

(a) Use the standard results for summations to show that

∑r=1n(2r−1)2=n3(An2+B)\sum_{r=1}^{n} (2r - 1)^2 = \frac{n}{3}(An^2 + B)

where AA and BB are integers to be determined.

(5)

(b) Use the answer to part (a) to determine the value of

252+272+292+⋯+95225^2 + 27^2 + 29^2 + \cdots + 95^2

You must make your method clear.

(3)

中文翻译

(a) 使用求和标准公式证明

∑r=1n(2r−1)2=n3(An2+B)\sum_{r=1}^{n} (2r - 1)^2 = \frac{n}{3}(An^2 + B)

其中 AA 和 BB 是待确定的整数。

(5)

(b) 利用 (a) 的答案确定

252+272+292+⋯+95225^2 + 27^2 + 29^2 + \cdots + 95^2

的值。必须清楚说明方法。

(3)

解答

(a)

解法一

思路

展开

先展开 (2r−1)2(2r-1)^2，把原式拆成 4∑r2−4∑r+∑14\sum r^2-4\sum r+\sum1。代入标准求和公式并提取公因式 n3\frac n3，最后与题目指定形式比较。

答题过程

展开

Using the standard results for summations,

∑r=1n(2r−1)2= 4∑r=1nr2−4∑r=1nr+∑r=1n1= 4[n(n+1)(2n+1)6] −4[n(n+1)2]+n= n3[2(n+1)(2n+1)−6(n+1)+3]= n3(4n2+6n+2−6n−6+3)= n3(4n2−1).\begin{align*} \sum_{r=1}^{n}(2r-1)^2 =&\,4\sum_{r=1}^{n}r^2 -4\sum_{r=1}^{n}r +\sum_{r=1}^{n}1\\ =&\,4\left[\frac{n(n+1)(2n+1)}{6}\right]\\ &\,-4\left[\frac{n(n+1)}{2}\right]+n\\ =&\,\frac{n}{3} \big[2(n+1)(2n+1)-6(n+1)+3\big]\\ =&\,\frac{n}{3} \big(4n^2+6n+2-6n-6+3\big)\\ =&\,\frac{n}{3}(4n^2-1). \end{align*}

Therefore, comparing with n3(An2+B)\dfrac n3(An^2+B),

A=4,B=−1.\boxed{A=4,\qquad B=-1}.

(b)

解法一

思路

展开

通项中的奇数是 2r−12r-1。解 2r−1=252r-1=25 与 2r−1=952r-1=95，确定所求范围是第 13 项至第 48 项，再用“前 48 项之和减去前 12 项之和”。

答题过程

展开

Since

25=2(13)−125=2(13)-1

and

95=2(48)−1,95=2(48)-1,

the required sum consists of the 13th to the 48th odd squares. Using part (a),

252+272+⋯+952= ∑r=148(2r−1)2−∑r=112(2r−1)2= 483[4(48)2−1] −123[4(12)2−1]= 16(9215)−4(575)= 147440−2300= 145140.\begin{align*} 25^2+27^2+\cdots+95^2 =&\,\sum_{r=1}^{48}(2r-1)^2 -\sum_{r=1}^{12}(2r-1)^2\\ =&\,\frac{48}{3}\big[4(48)^2-1\big]\\ &\,-\frac{12}{3}\big[4(12)^2-1\big]\\ =&\,16(9215)-4(575)\\ =&\,147440-2300\\ =&\,\boxed{145140}. \end{align*}