Series: arithmetic and geometric progressions, sums, binomial expansion

Series (Syllabus 1.6) – Arithmetic & Geometric Progressions, Binomial Expansion

Syllabus Checklist (Cambridge AS & A‑Level Mathematics 9709, 2026‑27)
  • Define arithmetic progression (AP) and geometric progression (GP).
  • Write the $n^{\text{th}}$ term of an AP and a GP.
  • Derive and use the formula for the sum of the first $n$ terms of an AP and a GP (finite).
  • State and apply the special cases $d=0$, $r=1$, $a=0$.
  • State the convergence condition $|r|<1$ for an infinite GP and calculate its sum to infinity.
  • Use $\Sigma$‑notation to express series and to derive the sum formulae.
  • State the binomial theorem for a positive integer exponent, write the general term, and read coefficients from Pascal’s triangle.

1. Arithmetic Progression (AP)

1.1 Definition & $n^{\text{th}}$ term

  • Sequence in which the difference between consecutive terms is constant – the common difference $d$.
  • First term: $a$.
  • General term: $t_n = a + (n-1)d$.

1.2 Sum of the first $n$ terms

Using $\Sigma$‑notation:

$$S_n = \sum_{k=1}^{n} t_k = \sum_{k=1}^{n}\bigl[a+(k-1)d\bigr]$$

Evaluating the arithmetic series (pairing the first and last terms) gives two equivalent closed‑form expressions:

$$S_n = \frac{n}{2}\bigl(2a+(n-1)d\bigr)\;=\;\frac{n}{2}\bigl(t_1+t_n\bigr)$$

1.3 Special cases

  • $d=0$ → All terms equal $a$; $S_n = na$.
  • $a=0$ → $ t_n = (n-1)d$ (the progression starts from zero).
  • $d<0$ → The AP decreases; the formulae remain valid.

1.4 Example

Find the sum of the first 15 terms of the AP $4,\;9,\;14,\dots$.

  1. Identify $a=4$, $d=5$.
  2. Find the 15‑th term: $t_{15}=a+(15-1)d=4+14\cdot5=74$.
  3. Apply the sum formula:
$$S_{15}= \frac{15}{2}\bigl(4+74\bigr)=\frac{15}{2}\times78=585.$$

2. Geometric Progression (GP)

2.1 Definition & $n^{\text{th}}$ term

  • Sequence in which each term after the first is obtained by multiplying the preceding term by a constant – the common ratio $r$.
  • First term: $a$.
  • General term: $g_n = a\,r^{\,n-1}$.

2.2 Sum of the first $n$ terms (finite GP)

In $\Sigma$‑notation:

$$G_n = \sum_{k=1}^{n} a\,r^{\,k-1}=a\sum_{k=0}^{n-1} r^{k}$$

Evaluating the geometric series (for $req1$) gives:

$$G_n = a\,\frac{1-r^{\,n}}{1-r}\qquad(req1)$$

2.3 Special cases

  • $r=1$ → All terms equal $a$; $G_n = na$.
  • $a=0$ → Every term is $0$; $G_n = 0$.
  • $r=-1$ → Terms alternate $a,-a,a,\dots$; use the same formula (note the sign of $r^{\,n}$).

2.4 Infinite GP – convergence

An infinite geometric series

$$\displaystyle\sum_{k=1}^{\infty} a\,r^{\,k-1}$$

converges **iff** $|r|<1$. When it converges, the sum to infinity is

$$G_{\infty}= \frac{a}{1-r}\qquad(|r|<1).$$

If $|r|\ge 1$ the series diverges (no finite sum).

2.5 Example (finite GP)

Find the sum of the first 6 terms of the GP $3,6,12,\dots$.

  1. $a=3$, $r=2$.
  2. Apply the finite‑sum formula:
$$G_6 = 3\,\frac{1-2^{6}}{1-2}=3\,\frac{1-64}{-1}=3\times63=189.$$

2.6 Example (infinite GP)

Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{2}{3^{\,n}}$.

  • Rewrite as $a r^{\,n-1}$ with $a=\frac{2}{3}$ and $r=\frac13$.
  • Since $|r|<1$, the series converges:
$$G_{\infty}= \frac{a}{1-r}= \frac{\frac{2}{3}}{1-\frac13}= \frac{\frac{2}{3}}{\frac23}=1.$$

3. Binomial Expansion

3.1 Binomial theorem (positive integer exponent)

For any non‑negative integer $n$:

$$ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k}\,x^{\,n-k}y^{\,k} $$

where the binomial coefficient

$$\displaystyle\binom{n}{k}= \frac{n!}{k!\,(n-k)!},\qquad 0\le k\le n.$$

3.2 General term

The $(k+1)^{\text{th}}$ term (counting from $k=0$) is

$$T_{k+1}= \binom{n}{k}\,x^{\,n-k}y^{\,k}.$$

3.3 Pascal’s triangle

Coefficients $\binom{n}{k}$ are read directly from Pascal’s triangle. The first six rows are shown below.

nCoefficients $\displaystyle\binom{n}{k}$
01
11 1
21 2 1
31 3 3 1
41 4 6 4 1
51 5 10 10 5 1

3.4 Example – expansion up to a specific term

Expand $(2x-3)^4$ up to the term in $x^2$.

  1. Identify $x\to2x$, $y\to-3$, $n=4$.
  2. Write the first three terms ($k=0,1,2$):
\[ \begin{aligned} (2x-3)^4 &= \binom{4}{0}(2x)^4(-3)^0 + \binom{4}{1}(2x)^3(-3)^1 + \binom{4}{2}(2x)^2(-3)^2 + \dots\\[4pt] &= 1\cdot16x^4 -4\cdot8x^3\cdot3 +6\cdot4x^2\cdot9 + \dots\\[4pt] &= 16x^4 -96x^3 +216x^2 +\dots \end{aligned} \]

4. Practice Questions

  1. Find the $20^{\text{th}}$ term of the AP $7,\,12,\,17,\dots$.
  2. Calculate the sum of the first 10 terms of the GP $5,\,15,\,45,\dots$.
  3. Determine whether the infinite series $\displaystyle\sum_{n=1}^{\infty}\frac{2}{3^{\,n}}$ converges, and if so, find its sum.
  4. Expand $(x+2)^5$ and state the coefficient of $x^3$.
  5. Given that the sum of the first $n$ terms of an AP is $S_n = 3n^2 + 5n$, find the first term $a$ and the common difference $d$.

5. Summary of Key Formulae

Series type General term Sum of $n$ terms Special notes
Arithmetic progression (AP) $t_n = a + (n-1)d$ $S_n = \displaystyle\frac{n}{2}\bigl(2a+(n-1)d\bigr)=\displaystyle\frac{n}{2}(t_1+t_n)$ Linear growth; $d$ may be positive, negative or zero.
Geometric progression (GP) $g_n = a\,r^{\,n-1}$ $G_n = a\,\displaystyle\frac{1-r^{\,n}}{1-r}\;(req1)$
$G_{\infty}= \displaystyle\frac{a}{1-r}\quad(|r|<1)$
Infinite sum exists only for $|r|<1$. $r=1$ gives $G_n=na$.
Binomial expansion $T_{k+1}= \displaystyle\binom{n}{k}x^{\,n-k}y^{\,k}$ $(x+y)^n = \displaystyle\sum_{k=0}^{n}\binom{n}{k}x^{\,n-k}y^{\,k}$ Coefficients read from Pascal’s triangle; $n$ must be a non‑negative integer.
Suggested diagram: a number line showing an arithmetic progression (equal spacing $d$) alongside a geometric progression (spacing multiplied by $r$).

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