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IAL 2025 May FP1 Q8

A Level / Edexcel / FP1

IAL 2025 May Paper · Question 8

Question

[!problem]

(a) Using the standard summation formulae show that

r=1n4r2=2n(n+1)(an+b)3\sum_{r=1}^{n} 4r^2 = \frac{2n(n+1)(an+b)}{3}

where aa and bb are integers to be determined.

(5)

(b) Prove by induction that, for nZ+n \in \mathbb{Z}^+

r=1nr(3r+2)=n2(n+1)(9n+7)4\sum_{r=1}^{n} r(3r+2) = \frac{n^2(n+1)(9n+7)}{4}

(5)

Using the results from parts (a) and (b) and showing all stages of your working,

(c) determine the value of nn for which

r=1nr(3r+2)=15r=12n(2r+1)\sum_{r=1}^{n} r(3r+2) = 15\sum_{r=1}^{2n} (2r+1)

(Solutions relying entirely on calculator technology are not acceptable.)

(3)

中文翻译

(a) 使用求和标准公式证明

r=1n4r2=2n(n+1)(an+b)3\sum_{r=1}^{n} 4r^2 = \frac{2n(n+1)(an+b)}{3}

其中 aabb 是待确定的整数。

(5)

(b) 用数学归纳法证明,对于 nZ+n \in \mathbb{Z}^+

r=1nr(3r+2)=n2(n+1)(9n+7)4\sum_{r=1}^{n} r(3r+2) = \frac{n^2(n+1)(9n+7)}{4}

(5)

使用 (a) 和 (b) 的结果,展示所有运算过程,

(c) 确定满足下式的 nn

r=1nr(3r+2)=15r=12n(2r+1)\sum_{r=1}^{n} r(3r+2) = 15\sum_{r=1}^{2n} (2r+1)

(完全依赖计算器技术的解答不可接受。)

(3)