📚 Master Geometric Series: Edexcel C2 June 2016 Past Paper Explained | 等比数列满分攻略：Edexcel C2 2016真题精讲
In the Edexcel A Level Mathematics exam, the geometric series is one of the most reliable topics you will meet. It sits inside the Core Mathematics C2 module, and Question 1 of the June 2016 C2 paper tested exactly this: applying the sum formula and deriving a term. This article breaks that question down step by step, then widens out to the full set of geometric-series skills C2 can examine, plus the exam technique and revision strategy that turn a good attempt into full marks.
在 Edexcel ALEVEL 数学考试中，等比数列（Geometric Series）是 Core Mathematics C2 模块的必考题型。2016 年 6 月的 C2 真题第一题就考察了等比数列的求和与通项推导。本文将逐步骤拆解这道真题，再扩展到 C2 可能考到的全部等比数列技能，并配上把"会做"变成"满分"的考试技巧与备考策略。
1. Why Geometric Series Is Guaranteed in C2 | 为什么等比数列在 C2 中必考
C2 carries 75 marks in 90 minutes, and a geometric-series question has appeared on virtually every sitting for the past decade. It usually arrives as Question 1 or 2, which means it is deliberately designed as a warm-up: the method is standard, the algebra is short, and the marks are there to be banked. Losing marks here hurts disproportionately, because the later questions on the paper are harder to recover from.
C2 试卷满分 75 分、考试时间 90 分钟，而过去十年里几乎每一场考试都出现了等比数列题。它通常出现在第 1 题或第 2 题，是命题人刻意安排的"热身题"：方法固定、运算简短、分数就是送分的。这里丢分会造成不成比例的损失，因为后面的大题更难把分数补回来。
The June 2016 paper opened with a reverse-engineering problem: candidates were given the common ratio and the sum of the first four terms, and had to work backwards to prove the value of the first term. That single question compresses three testable skills into one part - knowing the sum formula, substituting fractions correctly, and presenting a clean algebraic proof.
2016 年 6 月的试卷以一道"反向推导"题开场：题目给出公比和前四项之和，要求考生反推出首项。这一问把三项可考技能压缩在一起——记住求和公式、正确代入分数、以及写出干净的代数证明过程。
2. Core Formula 1: The nth Term | 核心公式一：通项公式
Every geometric series is defined by two numbers: the first term a and the common ratio r. Each term is obtained by multiplying the previous term by r, so the nth term is reached after n − 1 multiplications.
每一个等比数列都由两个数决定：首项 a 和公比 r。每一项都由前一项乘以 r 得到，因此第 n 项相当于做了 n − 1 次乘法。
uₙ = a·rⁿ⁻¹
From this single line you can derive everything else the examiner asks for. If you know any two of a, r and a specific term, you can find the third. A very common C2 command is "the fifth term is 20.25" or "the third term is 36, the sixth term is 4.5", both of which are solved by forming simultaneous equations from uₙ = a·rⁿ⁻¹.
由这一行可以推导出考官要问的几乎所有内容。只要知道 a、r 和某一项这三个量中的任意两个，就能求出第三个。C2 常见问法是"第五项是 20.25"或"第三项是 36，第六项是 4.5"，两者都通过 uₙ = a·rⁿ⁻¹ 联立方程求解。
For a ratio from two terms, divide the larger-index term by the smaller-index one so that a cancels:
若由两项求公比，就用指数较大的项除以指数较小的项，使 a 约掉：
uₘ ÷ uₖ = r⁽ᵐ⁻ᵏ⁾
Remember that r can be negative or fractional. A negative ratio makes the terms alternate in sign, which is a favourite trap in sign-handling questions.
要记住 r 可以是负数或分数。公比为负时数列符号交替出现，这在"符号处理"类题目中是常见陷阱。
3. Core Formula 2: Sum of the First n Terms | 核心公式二：前 n 项和公式
The sum of the first n terms has two equivalent forms, and choosing the right one saves time. When r ≠ 1:
前 n 项之和有两种等价形式，选对形式可以省时间。当 r ≠ 1 时：
Sₙ = a(1 − rⁿ) ÷ (1 − r) = a(rⁿ − 1) ÷ (r − 1)
The first version is easier when r < 1, because both the numerator bracket and the denominator stay positive. The second is easier when r > 1, for the same reason. The special case r = 1 gives a constant series, where Sₙ = na - do not forget it, because "explain why the formula breaks down when r = 1" is a legitimate one-mark question.
当 r < 1 时第一种更好用，因为分子括号和分母都保持正数；当 r > 1 时第二种更好用，道理相同。特殊情况 r = 1 对应常数数列，此时 Sₙ = na——不要忘记这一条，因为"解释 r = 1 时公式为何失效"是合理的一分小题。
Edexcel formula booklets do provide these results in some specifications, but C2 candidates are expected to know them instantly. Every extra second spent hunting through a booklet is a second lost from a later question.
部分考纲的公式手册中会给出这些公式，但 C2 考生应能做到"条件反射"。在公式表里翻找多花的每一秒，都是从后面题目里借来的时间。
4. June 2016 C2 Question 1: Step-by-Step | 2016 年 6 月 C2 第一题逐步精讲
The June 2016 C2 paper opened with a geometric series whose first term is a and whose common ratio is r didates were told that the sum of the first four terms is 175, and were asked to show that a = 64. Here is the full method.
2016 年 6 月 C2 试卷的第一题是一个等比数列，首项为 a，公比 r = 3/4。题目给出前四项之和为 175，要求证明 a = 64。完整步骤如下。
Step 1 - Write down the correct formula. ÷ (1 − r) with n = 4. Writing the formula earns the method mark even if arithmetic slips.
第一步——写出正确公式。 写出 Sₙ = a(1 − rⁿ) ÷ (1 − r)，并代入 n = 4。即使后面算错，写出公式也能拿到方法分。
Step 2 - Substitute the given values. Put r = 3/4 and S₄ = 175 into the formula:
第二步——代入已知值。 把 r = 3/4 和 S₄ = 175 代入公式：
175 = a(1 − (3/4)⁴) ÷ (1 − 3/4)
Step 3 - Evaluate the powers and the denominator. 1 − 81/256 = 175/256. Also 1 − 3/4 = 1/4. Dividing by 1/4 is the same as multiplying by 4.
第三步——算出幂与分母。 (3/4)⁴ = 81/256，所以 1 − 81/256 = 175/256；同时 1 − 3/4 = 1/4。除以 1/4 等价于乘以 4。
175 = a × (175/256) × 4 = 175a ÷ 64
Step 4 - Solve for a. Multiply both sides by 64 and divide by 175:
第四步——解出 a。 两边同乘 64，再同除 175：
175 × 64 = 175a ⟹ a = 64 ∎
The elegance of this question is that the factor 175 appears on both sides, so the arithmetic collapses neatly. If your final line is messier than that, you have almost certainly made an error earlier.
这道题的精妙之处在于 175 这个因子在两边同时出现，运算因此干净地收敛。如果你的最后一步比这更乱，几乎可以肯定前面某处出错了。
5. Verification and Reverse-Checking | 验证与反向检查
Once you have a = 64 and r ₃ = 36, u₄ = 27. Their sum is 64 + 48 + 36 + 27 = 175, which matches the question exactly.
得到 a = 64、r = 3/4 之后，花二十秒检验：写出前四项 u₁ = 64、u₂ = 48、u₃ = 36、u₄ = 27，它们的和是 64 + 48 + 36 + 27 = 175，与题目完全吻合。
This kind of direct check works whenever n is small, and it is worth the time because it catches a sign error or an inverted fraction instantly. For larger n, use a structural check instead: the ratio of consecutive terms must be constant, and the sum must lie between the first term and a ÷ (1 − r).
只要 n 较小，这种直接检验都适用，而且非常值得花时间，因为它能立刻发现符号错误或分数颠倒。n 较大时改用结构性检查：相邻项之比必须恒定，而且和必须落在首项与 a ÷ (1 − r) 之间。
In "show that" questions the marks are awarded for the chain of reasoning, not for the final value. Write every substitution on its own line, use exact fractions rather than decimals, and finish with the statement the question asked you to prove.
在"证明题（show that）"中，分数给的是推理链条，而不是最终数值。每一步代入单独写一行，使用精确分数而不是小数，最后写出题目要求你证明的结论。
6. Sum to Infinity and the Convergence Condition | 无穷等比级数与收敛条件
When the common ratio satisfies |r| < 1, the terms shrink towards zero and the series converges. Its sum to infinity is a clean, finite value:
当公比满足 |r| < 1 时，各项逐渐趋近于零，级数收敛，其无穷和是一个干净的有限值：
S∞ = a ÷ (1 − r) (valid only when |r| < 1)
6. Notice that 256 is greater than any finite Sₙ but still finite - a useful sanity check.
对 2016 年 6 月这道题中的数列，a = 64、r = 3/4，可得 S∞ = 64 ÷ (1 − 3/4) = 64 ÷ (1/4) = 256。注意 256 大于任何有限的 Sₙ，但仍然有限——这是很有用的合理性检验。
This topic is high-frequency in C2 and usually appears in the middle or later part of the paper, where it is paired with an inequality. Typical demands are: "state the condition for the series to converge", "find the sum to infinity", or "find the range of values of k for which the series converges", where the ratio is something like (k − 3) ÷ 2 and you must solve |(k − 3) ÷ 2| < 1.
这是 C2 的高频考点，通常出现在试卷中后段，并与不等式结合。常见要求包括："写出级数收敛的条件"、"求无穷和"、或"求 k 的取值范围使级数收敛"——此时公比形如 (k − 3) ÷ 2，需要解 |(k − 3) ÷ 2| < 1。
ating the condition explicitly is worth a mark. Do not simply write the value of S∞; write "since |r| = 3/4 < 1, the series converges and S∞ = ...".
明确写出收敛条件本身就有分。不要只写 S∞ 的数值，而应写成："由于 |r| = 3/4 < 1，级数收敛，故 S∞ = ..."。
7. Reverse Problems: Finding a and r from Sₙ | 逆向问题：由 Sₙ 反求 a 与 r
The June 2016 question is a reverse problem in its simplest form: Sₙ and r were given, and a was unknown. Harder versions give you two sums, or a sum and a term, and expect simultaneous equations.
2016 年 6 月这道题属于最简单的逆向问题：已知 Sₙ 和 r，求未知的 a。更难的版本会给出两个和，或一个和加一项，需要联立方程。
Given S₃ and S₆: form two equations, then divide one by the other to eliminate a and produce a polynomial in r.
已知 S₃ 与 S₆：列出两个方程，两式相除消去 a，得到关于 r 的多项式。
Given S∞ and a: rearrange S∞ = a ÷ (1 − r) to get r = 1 − a ÷ S∞, which is often the fastest route.
已知 S∞ 与 a：由 S∞ = a ÷ (1 − r) 变形得 r = 1 − a ÷ S∞，这通常是最快的路径。
and a sum: substitute uₙ = a·rⁿ⁻¹ into the sum equation and solve the resulting equation.
已知某一项与某个和：把 uₙ = a·rⁿ⁻¹ 代入求和方程，解所得方程。
Whenever you eliminate a, watch for r s surprisingly often, and it must either be accepted or explicitly rejected with a reason.
每次消去 a 时都要留意 r = 1 这个可能解。它出现得比想象中频繁，而你必须要么接受它，要么给出明确理由排除它。
8. Combining with Logarithms to Find the Least n | 与对数结合求最小 n
hold. With a = 64 and r = 3/4, suppose the question asks for the least n such that uₙ < 1.
一种常见的延伸题会问：从哪一项起数列的值小于某个阈值。以 a = 64、r = 3/4 为例，设题目要求满足 uₙ < 1 的最小 n。
Set up the inequality and take logarithms of both sides:
列出不等式并在两边取对数：
64 × (3/4)ⁿ⁻¹ < 1 ⟹ (n − 1)·ln(3/4) < ln(1/64)
Because ln(3/4) is negative, dividing by it reverses the inequality sign - this is the single most common error in the whole topic:
由于 ln(3/4) 为负，除以它时不等号方向要反转——这是整个知识点中最常见的错误：
n − 1 > ln(1/64) ÷ ln(3/4) ≈ 14.46 ⟹ n > 15.46 ⟹ n = 16
Always interpret the answer in context - if the question asks for a term number, you cannot leave it as a decimal.
检验一下：u₁₅ ≈ 1.14，u₁₆ ≈ 0.855，所以 n = 16 确实是第一个小于 1 的项。答案一定要回到题目语境中解释——如果问的是项数，就不能保留为小数。
9. Real-World Applications: Interest, Growth and Decay | 实际应用：复利、人口与衰减
Geometric series model any quantity that changes by a fixed percentage in each period. Compound interest, population growth and radioactive decay are all the same mathematics wearing different clothes.
等比数列刻画的是任何"每期按固定百分比变化"的量。复利、人口增长和放射衰变本质上都是同一套数学，只是换了外衣。
). In depreciation or radioactive decay, r < 1 and the model is the same. Edexcel frequently embeds that simple model inside a word problem and adds a second layer: "after how many years does the value first exceed £20 000?" - which is again a logarithm question.
在复利中，n 年后的价值为 a·rⁿ，其中 r = 1 + (年利率 ÷ 100)。在折旧或放射衰变中 r < 1，模型形式完全一样。Edexcel 常把这个简单模型包进应用题，再加入第二层要求："第几年价值首次超过 20000 英镑？"——这又回到对数问题。
When reading an application, extract three things before you calculate: the initial value (a), the multiplier per period (r), and the number of periods (n). Modelling marks are awarded for these identifications, so write them down explicitly.
读应用题时，先提取三件事再动笔：初始值（a）、每期乘数（r）、期数（n）。建模分会给在这些识别步骤上，所以务必明确写出来。
10. Exam Technique: Marks, Timing and Method Marks | 考试技巧：分值、时间与过程分
C2 is worth 75 marks in 90 minutes, which is roughly 1.2 minutes per mark. The numbers in brackets on the paper tell you exactly how much working is expected, so use them as a time budget. A four-mark part deserves about five minutes and a line of justification per mark.
C2 满分 75 分、时间 90 分钟，平均约每分 1.2 分钟。试卷上括号里的分值直接告诉你该写多少过程，把它当成时间预算：4 分的小问大约值 5 分钟，平均每一分对应一行论证。
Q1 on geometric series is a warm-up - secure full marks here before moving on.
第 1 题的等比数列是热身题——先在这里拿满分，再往后走。
Always show your working; an answer with no method can lose every mark except the final one.
务必写出过程；只有答案没有方法，除了最后一分外可能全部丢分。
Use exact fractions (3/4, 81/256) until the final line, then round only if the question asks for a decimal.
到最后一步之前都用精确分数（3/4、81/256），只有在题目要求小数时才四舍五入。
Never round intermediate values - premature rounding is the single biggest source of lost accuracy marks.
绝不要对中间值取近似——过早舍入是丢失精度分最大的来源。
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