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Series and Sequences

Subject: Additional Mathematics
Topic: 9
Cambridge Code: 4037 / 0606


Sequences​

Sequence - Ordered list of numbers following a pattern

Notation​

a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_n

where:

  • ana_n is the nn-th term
  • a1a_1 is the first term

General Term (Explicit Formula)​

Formula for the nn-th term: ana_n or TnT_n

Example: For sequence 3, 6, 9, 12, ... an=3na_n = 3n


Arithmetic Sequences​

Arithmetic Sequence - Sequence with constant difference between consecutive terms

Common Difference​

d=an+1−and = a_{n+1} - a_n

General Formula​

an=a1+(n−1)da_n = a_1 + (n-1)d

where:

  • ana_n is the nn-th term
  • a1a_1 is the first term
  • dd is the common difference

Example​

Sequence: 2, 5, 8, 11, ...

  • First term: a1=2a_1 = 2
  • Common difference: d=3d = 3
  • General formula: an=2+(n−1)(3)=3n−1a_n = 2 + (n-1)(3) = 3n - 1
  • 10th term: a10=3(10)−1=29a_{10} = 3(10) - 1 = 29

Sum of Arithmetic Series​

Sn=n2(a1+an)=n2(2a1+(n−1)d)S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}(2a_1 + (n-1)d)

where SnS_n is the sum of first nn terms

Example​

Sum of first 10 terms of 2, 5, 8, 11, ...:

S10=102(2+29)=5×31=155S_{10} = \frac{10}{2}(2 + 29) = 5 \times 31 = 155

Or: S10=102(2(2)+9(3))=5(4+27)=155S_{10} = \frac{10}{2}(2(2) + 9(3)) = 5(4 + 27) = 155


Geometric Sequences​

Geometric Sequence - Sequence with constant ratio between consecutive terms

Common Ratio​

r=an+1anr = \frac{a_{n+1}}{a_n}

General Formula​

an=a1⋅rn−1a_n = a_1 \cdot r^{n-1}

where:

  • ana_n is the nn-th term
  • a1a_1 is the first term
  • rr is the common ratio

Example​

Sequence: 3, 6, 12, 24, ...

  • First term: a1=3a_1 = 3
  • Common ratio: r=2r = 2
  • General formula: an=3⋅2n−1a_n = 3 \cdot 2^{n-1}
  • 6th term: a6=3⋅25=3⋅32=96a_6 = 3 \cdot 2^5 = 3 \cdot 32 = 96

Sum of Geometric Series​

If r≠1r \neq 1: Sn=a1⋅rn−1r−1=a1⋅1−rn1−rS_n = a_1 \cdot \frac{r^n - 1}{r - 1} = a_1 \cdot \frac{1 - r^n}{1 - r}

If r=1r = 1: Sn=n⋅a1S_n = n \cdot a_1

Example​

Sum of first 6 terms of 3, 6, 12, 24, ...:

S6=3⋅26−12−1=3⋅64−11=3⋅63=189S_6 = 3 \cdot \frac{2^6 - 1}{2 - 1} = 3 \cdot \frac{64 - 1}{1} = 3 \cdot 63 = 189


Infinite Geometric Series​

Convergent Series - Series with a finite sum (when ∣r∣<1|r| < 1)

Sum to Infinity​

S∞=a11−r,∣r∣<1S_\infty = \frac{a_1}{1 - r}, \quad |r| < 1

Example​

Sum of infinite series 2+1+0.5+0.25+…2 + 1 + 0.5 + 0.25 + \ldots

  • First term: a1=2a_1 = 2
  • Common ratio: r=0.5r = 0.5

S∞=21−0.5=20.5=4S_\infty = \frac{2}{1 - 0.5} = \frac{2}{0.5} = 4

Divergent Series: If ∣r∣≥1|r| \geq 1, series diverges (no finite sum)


Sigma Notation​

Sigma Notation - Compact way to write a sum

∑n=1Nan=a1+a2+a3+⋯+aN\sum_{n=1}^{N} a_n = a_1 + a_2 + a_3 + \cdots + a_N

Examples​

∑n=152n=2(1)+2(2)+2(3)+2(4)+2(5)=2+4+6+8+10=30\sum_{n=1}^{5} 2n = 2(1) + 2(2) + 2(3) + 2(4) + 2(5) = 2+4+6+8+10 = 30

∑n=143n=31+32+33+34=3+9+27+81=120\sum_{n=1}^{4} 3^n = 3^1 + 3^2 + 3^3 + 3^4 = 3+9+27+81 = 120


Sum Formulas​

Sum of Natural Numbers​

∑n=1Nn=N(N+1)2\sum_{n=1}^{N} n = \frac{N(N+1)}{2}

Sum of Squares​

∑n=1Nn2=N(N+1)(2N+1)6\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}

Sum of Cubes​

∑n=1Nn3=[N(N+1)2]2\sum_{n=1}^{N} n^3 = \left[\frac{N(N+1)}{2}\right]^2


Fibonacci and Other Sequences​

Fibonacci Sequence: Fn=Fn−1+Fn−2F_n = F_{n-1} + F_{n-2}

Starting with F1=1,F2=1F_1 = 1, F_2 = 1: 1, 1, 2, 3, 5, 8, 13, 21, ...

Recursive Sequences​

Defined by a recurrence relation: an=f(an−1,an−2,…)a_n = f(a_{n-1}, a_{n-2}, \ldots)

Example: an=2an−1+1a_n = 2a_{n-1} + 1 with a1=1a_1 = 1

  • a1=1a_1 = 1
  • a2=2(1)+1=3a_2 = 2(1) + 1 = 3
  • a3=2(3)+1=7a_3 = 2(3) + 1 = 7
  • a4=2(7)+1=15a_4 = 2(7) + 1 = 15

Key Points to Remember​

  1. Arithmetic: Constant difference dd
  2. Geometric: Constant ratio rr
  3. Arithmetic sum: Sn=n2(a1+an)S_n = \frac{n}{2}(a_1 + a_n)
  4. Geometric sum: Sn=a1⋅1−rn1−rS_n = a_1 \cdot \frac{1-r^n}{1-r}
  5. Infinite geometric sum: S∞=a11−rS_\infty = \frac{a_1}{1-r} (if ∣r∣<1|r| < 1)
  6. Use sigma notation for compact representation

Worked Examples​

Example 1: Arithmetic Sequence​

An arithmetic sequence has a3=7a_3 = 7 and a7=15a_7 = 15. Find a1a_1 and dd:

From an=a1+(n−1)da_n = a_1 + (n-1)d:

  • a3:7=a1+2da_3: 7 = a_1 + 2d ... (1)
  • a7:15=a1+6da_7: 15 = a_1 + 6d ... (2)

Subtract (1) from (2): 8=4d⇒d=28 = 4d \Rightarrow d = 2

From (1): 7=a1+4⇒a1=37 = a_1 + 4 \Rightarrow a_1 = 3

Example 2: Geometric Sum​

A geometric series has a1=2a_1 = 2 and r=12r = \frac{1}{2}. Find S5S_5:

S5=2⋅1−(12)51−12=2⋅1−13212=2⋅2⋅3132=318=3.875S_5 = 2 \cdot \frac{1 - (\frac{1}{2})^5}{1 - \frac{1}{2}} = 2 \cdot \frac{1 - \frac{1}{32}}{\frac{1}{2}} = 2 \cdot 2 \cdot \frac{31}{32} = \frac{31}{8} = 3.875

Example 3: Infinite Series​

Find sum of infinite geometric series: 12+14+18+…\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots

  • a1=12a_1 = \frac{1}{2}, r=12r = \frac{1}{2}

S∞=121−12=1212=1S_\infty = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1


Practice Questions​

  1. Find the general term and 20th term of: 5, 10, 15, 20, ...

  2. Find the sum of first 8 terms of: 3, 9, 27, 81, ...

  3. Find the sum to infinity of: 1+12+14+18+…1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots

  4. Given arithmetic sequence with a2=5a_2 = 5 and a5=14a_5 = 14, find a10a_{10}.


Revision Tips​

  • Arithmetic: common difference dd (add/subtract)
  • Geometric: common ratio rr (multiply/divide)
  • Two equations needed to find a1a_1 and dd
  • Check if geometric series converges before finding S∞S_\infty
  • Use sigma notation for multiple terms