Sequences and series · Suites et séries
| English | Français |
|---|---|
| sequence/ˈsiːkwəns/ | séquence |
| arithmetic sequence/əˈrɪθmətɪk ˈsiːkwəns/ | suite arithmétique |
| geometric sequence/ˌdʒiːəʊˈmetrɪk ˈsiːkwəns/ | suite géométrique |
| series/ˈsɪəriːz/ | série |
| common difference/ˈkɒmən ˈdɪfrəns/ | différence constante |
| common ratio/ˈkɒmən ˈreɪʃɪəʊ/ | rapport constant |
| sum to infinity/sʌm tʊ ɪnˈfɪnɪti/ | somme à l'infini |
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? · Un investissement augmente de 4 % chaque année. À quelle famille appartient-il ?
A percentage is a factor, not an amount, so each year multiplies by 1.04. · Un pourcentage est un facteur, pas une quantité, donc chaque année on multiplie par 1,04.
Keep the term index explicit
- A sequence 数列 is an ordered list; a series · série 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 · Arithmétique contre géométrique
A constant factor eventually overtakes a constant amount, however large the amount. · Un facteur constant finit toujours par surpasser une quantité constante, quelle que soit l'importance de cette quantité.
An arithmetic sequence starts at 5 with common difference 3. What is the 10th term? · Une suite arithmétique commence à 5 avec une raison commune de 3. Quel est le 10e terme ?
u₁₀ = 5 + 9 × 3 = 32. Nine steps from the first term, not ten. · u₁₀ = 5 + 9 × 3 = 32. Neuf pas depuis le premier terme, pas dix.
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... · Pour un premier terme non nul, une suite géométrique admet une somme finie à l'infini uniquement lorsque...
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. · Pour a non nul, la convergence exige |r| < 1. Lorsque |r| ≥ 1, les termes ne tendent pas vers zéro ; les sommes partielles peuvent diverger ou osciller, plutôt que toujours croître vers l'infini positif.
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? · Une machine vaut 80000 perd 12% par an. Quelle est sa valeur après 5 ans, arrondi au 100?
80000 × 0.88⁵ = 42218.553344, which rounds to 42200 to the nearest 100. The retained proportion is multiplied five times. · 80000 × 0.88⁵ = 42218.553344, ce qui s'arrondit à 42200 au proche 100. La proportion retenue est multipliée cinq fois.
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. · Perdre 12 % par an pendant cinq ans ne signifie pas perdre 60 % au total.
0.88⁵ = 0.528, so about 47% is lost. Each year's loss applies to a smaller amount than the last. · 0,88⁵ = 0,528, donc environ 47 % est perdu. Chaque perte annuelle s'applique à une quantité plus petite que la précédente.