📚 IB Math: The Chessboard Wheat Problem and Geometric Sequences | IB数学：棋盘麦粒问题与等比数列
According to an ancient legend, when the inventor of chess presented the game to a king, the ruler was so pleased that he offered any reward. The inventor asked for wheat: one grain on the first square, two on the second, four on the third, and so on, doubling each square until all 64 squares were filled. The king laughed at such a modest request—until the royal mathematicians calculated the true total.
根据一个古老的传说，当国际象棋的发明者把棋献给国王时，国王非常高兴，愿意赐予任何奖赏。发明者请求得到小麦：第一个格子上放一粒，第二个格子放两粒，第三个格子放四粒，依此类推，每个格子翻倍直到填满64个格子。国王对这个看似谦卑的请求一笑置之——直到皇家数学家计算出真实总数为止。
1. The Legend | 传说
The chessboard wheat problem appears in many cultures, often with grains of rice or wheat. The key idea is that each square receives double the number of grains of the previous square. This creates a sequence of powers of two: 1, 2, 4, 8, 16, 32, 64, 128, and so on.
棋盘麦粒问题在许多文化中都有出现，有时用米粒或麦粒。核心思想是每个格子得到的粒数是前一个格子的两倍。这产生了一个2的幂构成的数列：1、2、4、8、16、32、64、128，如此类推。
The inventor’s request was not actually humble. The total amount of wheat would require a silo much larger than any kingdom could build. In modern terms, it would be hundreds of billions of tonnes, more than the entire world has produced in all of history combined.
发明者的请求其实并不谦卑。所需小麦的总量将需要一个比任何王国能建造的还要大得多的粮仓。用现代标准衡量，这将是数千亿吨，超过了全世界有史以来的小麦产量总和。
2. What Is a Geometric Sequence? | 什么是等比数列
A geometric sequence is a list of numbers in which each term after the first is obtained by multiplying the previous term by a fixed non-zero constant called the common ratio. For example: 3, 6, 12, 24 has common ratio 2.
等比数列是一列数字，其中从第一项以后，每一项都是把前一项乘以一个固定的非零常数得到的，这个常数称为公比。例如：3、6、12、24的公比为2。
For a sequence with first term a and common ratio r, the n-th term is given by the formula:
对于首项为a、公比为r的数列，其第n项由以下公式给出：
uₙ = a × rⁿ⁻¹
In the chessboard problem, a = 1 and r = 2. The number of grains on the 64th square is u₆₄ = 1 × 2⁶³, an astronomically large number.
在棋盘问题中，a = 1，r = 2。第64个格子上的粒数是 u₆₄ = 1 × 2⁶³，一个天文数字。
3. Modeling the Wheat Problem | 建模
To model the whole chessboard, we assign each square an index from 1 to 64. Square 1 has 1 grain, square 2 has 2 grains, square 3 has 4 grains, and square k has 2ᵏ⁻¹ grains. This is a geometric sequence with a = 1 and r = 2.
为了对整个棋盘建模，我们给每个格子编号1到64。第1格有1粒，第2格有2粒，第3格有4粒，第k格有2ᵏ⁻¹粒。这是一个 a = 1、r = 2 的等比数列。
The total number of grains is the sum of this sequence from square 1 to square 64. Rather than adding 64 numbers one by one, we use the geometric series formula.
总粒数是这个数列从第1格到第64格的和。与其一个一个地累加64个数字，不如使用等比数列求和公式。
Let us denote the sum of the first n terms as Sₙ. For the chessboard, n = 64, so we need S₆₄ = 1 + 2 + 4 + 8 + … + 2⁶³.
我们把前n项之和记为Sₙ。对于棋盘问题，n = 64，所以我们要求 S₆₄ = 1 + 2 + 4 + 8 + … + 2⁶³。
4. The Sum Formula | 求和公式
For a geometric sequence with first term a, common ratio r, and r ≠ 1, the sum of the first n terms is:
对于首项a、公比r（且 r ≠ 1）的等比数列，前n项之和为：
Sₙ = a(rⁿ − 1) / (r − 1)
Here is why the formula works. Write Sₙ = a + ar + ar² + … + arⁿ⁻¹. Multiply every term by r, then subtract the original sum. All middle terms cancel, leaving Sₙ(r − 1) = arⁿ − a. Rearranging gives the formula.
下面说明公式为何成立。写出 Sₙ = a + ar + ar² + … + arⁿ⁻¹。将每一项乘以r，再减去原来的和。所有中间项相互抵消，剩下 Sₙ(r − 1) = arⁿ − a。移项即可得到公式。
Notice that the formula can also be written as Sₙ = a(1 − rⁿ) / (1 − r). Both forms are equivalent; choose whichever makes the denominator positive for your problem.
注意公式也可以写成 Sₙ = a(1 − rⁿ) / (1 − r)。两种形式等价；可以根据题目选择让分母为正的那种形式。
5. Bringing in the Chessboard Numbers | 代入棋盘数据
For the wheat problem, we have a = 1, r = 2, and n = 64. Substituting into the alternative form:
对麦粒问题，a = 1，r = 2，n = 64。代入另一种形式：
S₆₄ = 1(2⁶⁴ − 1) / (2 − 1) = 2⁶⁴ − 1 = 18,446,744,073,709,551,615
The total number of grains is 18 quintillion, 446 quadrillion, 744 trillion, 73 billion, 709 million, 551 thousand, 615. This number is often cited as 2⁶⁴ − 1.
总粒数是1844京6744兆0737亿0955万1615。这个数字常被写作2⁶⁴ − 1。
This value is famous in computer science as the maximum 64-bit unsigned integer, showing how closely mathematics connects to real-world technology.
这个值在计算机科学中很有名，它是64位无符号整数的最大值，体现了数学与现实技术之间的紧密联系。
6. Understanding Exponential Growth | 理解指数增长
Exponential growth means that the quantity grows by a fixed percentage or factor each step. Here the factor is 2. While the first few squares contain tiny amounts, the growth quickly becomes overwhelming.
指数增长是指每一步都以固定的百分比或倍数增长。这里的倍数是2。虽然前几个格子的数量很小，但增长很快变得惊人。
| Square | Grains on that square | Total so far |
| 10 | 512 | 1,023 |
| 20 | 524,288 | 1,048,575 |
| 30 | 536,870,912 | 1,073,741,823 |
| 40 | 549,755,813,888 | 1,099,511,627,775 |
| 64 | 9,223,372,036,854,775,808 | 18,446,744,073,709,551,615 |
Notice that the total sum up to any square is always one less than the grain count on the next square. This pattern is a beautiful consequence of the identity 2⁰ + 2¹ + 2² + … + 2ⁿ⁻¹ = 2ⁿ − 1.
可以看到，到任意格子为止的总和，总是比下一个格子的粒数少1。这个规律是恒等式 2⁰ + 2¹ + 2² + … + 2ⁿ⁻¹ = 2ⁿ − 1 的优美结果。
7. Large Numbers and Scientific Notation | 大数与科学计数法
The full number 2⁶⁴ − 1 is difficult to read and compare. In IB mathematics, expressing very large numbers in scientific notation helps. We can approximate 2⁶⁴ ≈ 1.845 × 10¹⁹. This approximation is useful for estimation and real-world comparison.
完整数字2⁶⁴ − 1难以读写和比较。在IB数学中，用科学计数法表达非常大的数字很有帮助。我们可以近似地写 2⁶⁴ ≈ 1.845 × 10¹⁹。这种近似对估算和现实比较非常有用。
If one tonne of wheat contains about 30 million grains, then 18.4 quintillion grains corresponds to roughly 6.15 × 10¹¹ tonnes. That is more than 600 billion tonnes—far more than an entire planet could store.
如果1吨小麦大约含有3000万粒麦粒，那么1844京粒大约相当于6.15 × 10¹¹吨。那就是超过6000亿吨——远远超过整个地球能够储存的数量。
Scientific notation also helps when checking answers. An IB exam question rarely requires writing out 20 digits; converting to standard form shows mathematical maturity and avoids errors.
科学计数法也有助于检查答案。IB考试题目很少要求写出20位数字；转换成标准形式显示了数学素养并避免错误。
8. Geometric Series in the IB Formula Booklet | IB公式手册中的等比级数
The IB Analysis and Approaches formula booklet provides two key results for geometric sequences. The n-th term is uₙ = u₁ · rⁿ⁻¹, and the sum of n terms is Sₙ = u₁(rⁿ − 1)/(r − 1). For infinite series with |r| < 1, the sum is S∞ = u₁/(1 − r).
IB分析与方法公式手册提供了关于等比数列的两个关键结果。第n项为 uₙ = u₁ · rⁿ⁻¹，n项和为 Sₙ = u₁(rⁿ − 1)/(r − 1)。对于 |r| < 1 的无穷级数，和为 S∞ = u₁/(1 − r)。
You may choose which letter to use, but in exams you should write both the formula and your substitution. For example, S₆₄ = 1(2⁶⁴ − 1)/(2 − 1) shows clear working.
你可以选择使用哪种字母，但在考试中应该写出公式和代入过程。例如，S₆₄ = 1(2⁶⁴ − 1)/(2 − 1) 展示了清晰的过程。
A common IB question asks you to find the number of terms n when the sum exceeds a certain value. This requires solving an inequality of the form a(rⁿ − 1)/(r − 1) > M, which can be done with logarithms.
一个常见的IB题目是当和超过某个数值时求项数n。这需要解形如 a(rⁿ − 1)/(r − 1) > M 的不等式，可以用对数来求解。
9. Common Pitfalls | 常见误区
One common mistake is confusing the number of terms with the last exponent. On a 64-square board, the last term has exponent 63, not 64, because the first term is 2⁰. The sum still uses r⁶⁴ in the numerator because the formula requires n terms.
一个常见错误是把项数与最后的指数混淆。在64个格子的棋盘上，最后一项的指数是63而不是64，因为第一项是2⁰。但求和公式的分子中仍然需要 r⁶⁴，因为公式要求有n项。
Another pitfall is using the infinite series formula when r > 1. The infinite geometric series only converges when |r| < 1. For the chessboard, r = 2, so no finite sum exists; the series diverges.
另一个误区是在 r > 1 时使用无穷级数公式。无穷等比级数只有在 |r| < 1 时才收敛。对于棋盘，r = 2，因此不存在有限和；该级数发散。
Students also forget that the first term may not be 1. If a problem starts with a different first term, always identify a and r explicitly before substituting into the formula.
学生有时会忘记首项不一定为1。如果题目给定了不同的首项，必须在代入公式之前明确找出a和r。
10. Worked Example and Exam-Style Question | 例题与考试风格题
Example: A geometric sequence has first term 3 and common ratio 4. Find the sum of the first 10 terms.
例题：一个等比数列的首项为3，公比为4。求前10项之和。
S₁₀ = 3(4¹⁰ − 1) / (4 − 1) = 3(1,048,576 − 1) / 3 = 1,048,575
Exam-style question: On the first day, a student answers 2 questions. Each following day, the student answers twice as many questions as the previous day. After 12 days, how many questions has the student answered in total?
考试风格题：第一天，一名学生做了2道题。此后每天做的题数是前一天的两倍。12天之后，该学生一共做了多少道题？
Here a = 2 and r = 2. The sum after 12 days is:
这里 a = 2，r = 2。12天后的总和为：
S₁₂ = 2(2¹² − 1) / (2 − 1) = 2 × 4095 = 8190
This problem illustrates how a simple doubling rule can generate large totals quickly—exactly the lesson of the chessboard.
这个问题说明了一个简单的翻倍规则如何能迅速产生很大的总数——这正是棋盘问题所要传达的道理。
11. Extension: Doubling and Other Ratios | 拓展：翻倍与其他比值
What if the common ratio were 3 instead of 2? Then the grains on the board would follow 1, 3, 9, 27, … and the total would be S₆₄ = (3⁶⁴ − 1)/2, which is even larger than 2⁶⁴ − 1. Exponential growth with a larger base accelerates even more dramatically.
如果公比不是2而是3呢？那么棋盘上的粒数就会是1、3、9、27、…，总数将是 S₆₄ = (3⁶⁴ − 1)/2，这比2⁶⁴ − 1还要大得多。底数更大的指数增长会更加迅猛。
If the common ratio were 0.5 instead, the sequence would be 1, 0.5, 0.25, 0.125, … and the total would approach 2 but never reach it. This is the infinite geometric series with sum S∞ = a/(1 − r) = 1/(1 − 0.5) = 2.
如果公比为0.5，数列就会是1、0.5、0.25、0.125、…，总和会趋近于2但永远达不到2。这就是无穷等比级数，其和为 S∞ = a/(1 − r) = 1/(1 − 0.5) = 2。
Understanding the contrast between r > 1 and |r| < 1 is fundamental to IB Mathematics: exponential growth versus exponential decay, divergent sums versus convergent sums.
理解 r > 1 与 |r| < 1 之间的对比是IB数学的基础：指数增长与指数衰减，发散之和与收敛之和。
12. The Bigger Picture | 更广阔的意义
The chessboard wheat problem is more than a classroom curiosity. It explains why exponential growth surprises us, why loans can spiral, why populations can explode, and why computers have limits on integer size.
棋盘麦粒问题不止是课堂上的趣闻。它解释了为什么指数增长让我们惊讶，为什么贷款会滚雪球，为什么人口会爆炸式增长，以及为什么计算机对整数的大小有限制。
In IB Mathematics, geometric sequences appear in financial mathematics, biological growth models, radioactive decay, and geometric series convergence tests. Mastering the formulas and their derivations gives you a toolkit for all these contexts.
在IB数学中，等比数列出现在金融数学、生物增长模型、放射性衰变以及几何级数收敛性判断等许多情境中。熟练掌握这些公式及其推导，会为你提供应对所有这些情境的工具箱。
So the next time you see a seemingly humble doubling question, remember the king’s grain: even the smallest beginning can grow beyond imagination.
所以，下次当你看到一个看似简单的翻倍问题时，请记住国王的麦粒：最微小的开端，也可能生长到超乎想象。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply