| English | 意味 |
|---|---|
| sequence/ˈsiːkwəns/ | 数列 |
| arithmetic sequence/əˈrɪθmətɪk ˈsiːkwəns/ | 等差数列 |
| geometric sequence/ˌdʒiːəʊˈmetrɪk ˈsiːkwəns/ | 等比数列 |
| series/ˈsɪəriːz/ | 級数 |
| common difference/ˈkɒmən ˈdɪfrəns/ | 公差 |
| common ratio/ˈkɒmən ˈreɪʃɪəʊ/ | 公比 |
| sum to infinity/sʌm tʊ ɪnˈfɪnɪti/ | 無限和 |
Match the rule to the sequence
- Saving a fixed amount each period gives an arithmetic sequence 等差数列 if nothing else changes. Multiplying the current amount by a fixed factor gives a geometric sequence 等比数列.
- A financial problem may include fixed payments, interest or both. Identify its stated rule rather than assuming every money problem is geometric.
An investment grows by 4% each year. Which family is it?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
A percentage is a factor, not an amount, so each year multiplies by 1.04.
Keep the term index explicit
- A sequence 数列 is an ordered list; a series 级数 adds its terms. Arithmetic uses common difference 公差 d and $u_n=a+(n-1)d$.
- Geometric uses common ratio 公比 r and $u_n=ar^{n-1}$. If the starting value is $u_1$, five completed multiplications give $u_6$; if a model uses time n starting at zero, write that convention instead.
Arithmetic against geometric
A constant factor eventually overtakes a constant amount, however large the amount.
An arithmetic sequence starts at 5 with common difference 3. What is the 10th term?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
u₁₀ = 5 + 9 × 3 = 32. Nine steps from the first term, not ten.
A sum is different from a term
- Arithmetic totals use $S_n=n[2a+(n-1)d]/2$. Geometric totals use $S_n=a(1-r^n)/(1-r)$ for $r\ne1$, or na when $r=1$.
- For nonzero a, a finite sum to infinity 无穷和 exists only when $|r|<1$, giving $S_\infty=a/(1-r)$. If a is zero, every term and sum is zero.
For a nonzero first term, a geometric series has a finite sum to infinity only when...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
For nonzero a, convergence requires |r| < 1. When |r| ≥ 1, terms do not tend to zero; partial sums may diverge or oscillate, rather than always growing to positive infinity.
A stated depreciation model. Starting value 80000 units loses 12% per year, with no other changes. $V(n)=V_0(1-r)^n$, so $V(5)=80000(0.88)^5=42218.553344$ units. This is 42200 to the nearest 100, not an exact value of 42200.
A machine worth 80000 loses 12% a year. What is it worth after 5 years, to the nearest 100?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
80000 × 0.88⁵ = 42218.553344, which rounds to 42200 to the nearest 100. The retained proportion is multiplied five times.
Use the sum for a combined capacity. Rows with counts 12, 14, 16 and so on give $u_{15}=40$, but $S_{15}=390$. Sheet 2.1 compares an individual term, a finite total and an infinite-horizon mathematical model.
For nonzero first term, ratios of magnitude at least 1 prevent convergence of the infinite geometric series. They do not all produce sums growing to positive infinity: ratio negative 1 makes partial sums alternate. State the convergence failure precisely.
Losing 12% a year for five years means losing 60% in total.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
0.88⁵ = 0.528, so about 47% is lost. Each year's loss applies to a smaller amount than the last.