Sequences and series

Definition. A sequence is arithmetic if every term differs from the previous by a constant, the common difference $d$. So $u_{n + 1} - u_{n} = d$.

$n$th term.

$u_{n} = a + (n - 1) d.$

Note: the first term $u_{1} = a$ — there are $(n - 1)$ common differences added, not $n$.

Sum to $n$ terms.

$S_{n} = \dfrac{n}{2}[2a + (n - 1) d] = \dfrac{n}{2}(a + l)$

where $l = u_{n}$ is the last term. Both forms are in the IAL formula booklet.

Derivation idea. Write the sum forward and backward, add them: each pair sums to $a + l$, and there are $n$ pairs (each counted twice). Hence $2 S_{n} = n(a + l)$.

Worked example. The $3$rd term is $11$ and the $8$th term is $26$. Find $S_{20}$.

• $u_{3} = a + 2d = 11$; $u_{8} = a + 7d = 26$.
• Subtract: $5d = 15$, $d = 3$. Then $a = 5$.
• $S_{20} = \tfrac{20}{2}[2(5) + 19(3)] = 10 \cdot 67 = 670$.

Modelling example. Savings: $50 in month 1, increasing by $5 each month. $a = 50$, $d = 5$. After $12$ months: $u_{12} = 50 + 11(5) = 105$ (so $105 in month 12). Total over $12$ months: $S_{12} = 6(100 + 55) = 930$.

Definition. A sequence is geometric if every term is a constant multiple of the previous — the common ratio $r$. So $u_{n + 1}/u_{n} = r$.

$n$th term.

$u_{n} = a r^{n - 1}.$

Note: power is $n - 1$, not $n$. The first term has $r^{0} = 1$.

Sum to $n$ terms.

$S_{n} = \dfrac{a(1 - r^{n})}{1 - r} = \dfrac{a(r^{n} - 1)}{r - 1} \quad (r \neq 1).$

Both forms are in the IAL formula booklet. Use whichever keeps the numerator positive — typically the first for $r < 1$, the second for $r > 1$.

Worked example. A geometric series has $a = 24$, $r = 3/4$. Find $S_{10}$.

• $S_{10} = \dfrac{24(1 - (3/4)^{10})}{1 - 3/4} = \dfrac{24(1 - 0.0563)}{0.25} = \dfrac{24 \cdot 0.9437}{0.25} \approx 90.60$.

Sum to infinity. When $|r| < 1$, the term $r^{n} \to 0$ as $n \to \infty$, and the series converges to:

$S_{\infty} = \dfrac{a}{1 - r} \quad (|r| < 1).$

(Given in the IAL formula booklet.) When $|r| \geq 1$ the series diverges and no sum to infinity exists.

Worked example. With $a = 24$, $r = 3/4$: $S_{\infty} = 24 / (1/4) = 96$. (And we saw $S_{10} \approx 90.60$, so the series approaches $96$ rapidly.)

Modelling: depreciation. A car worth $24000 depreciates by $15\%$ each year. After $n$ years: $V_{n} = 24000 \cdot 0.85^{n}$ (multiplier $0.85$, NOT $0.15$). After $5$ years: \approx \10649. Time to fall below \5000: solve $0.85^{n} < 5/24$, giving $n \geq 10$.

Notation.

$\sum_{n = m}^{N} f(n) = f(m) + f(m + 1) + \ldots + f(N).$

The variable $n$ below $\Sigma$ is the index of summation, running from the lower limit $m$ to the upper limit $N$. Number of terms: $N - m + 1$.

Strategy. Identify the sequence type from the formula $f(n)$, then apply the appropriate sum formula.

$f(n)$ has form	Sequence type
$A n + B$ (linear in $n$)	Arithmetic
$A \cdot b^{n - c}$ or $A b^{n}$	Geometric
$n^{2}, n^{3}, \ldots$	Standard results (not in P2)

Worked example. $\displaystyle \sum_{n = 1}^{20} (3n + 2)$.

• Arithmetic with first term $f(1) = 5$, last term $f(20) = 62$, $20$ terms.
• $S_{20} = \tfrac{20}{2}(5 + 62) = 670$.

Worked example. $\displaystyle \sum_{n = 1}^{8} 5 \cdot 2^{n - 1}$.

• Geometric with $a = 5$, $r = 2$, $8$ terms.
• $S_{8} = \dfrac{5(2^{8} - 1)}{2 - 1} = 5 \cdot 255 = 1275$.

Calculator note. Most graphing calculators have a $\sum$ button — use it to check your formula-based answer, but always show formula working for method marks.

Limits matter. $\displaystyle \sum_{n = 3}^{10} u_{n} = \sum_{n = 1}^{10} u_{n} - \sum_{n = 1}^{2} u_{n} = S_{10} - S_{2}$. So if the lower limit isn't $1$, compute the total minus the early-terms sum.

Definition. A recurrence relation defines a sequence using previous terms. Standard form: $u_{n + 1} = f(u_{n})$, with a stated initial value $u_{1}$.

Worked example. $u_{n + 1} = 2 u_{n} + 3$, $u_{1} = 1$.

• $u_{1} = 1$.
• $u_{2} = 2(1) + 3 = 5$.
• $u_{3} = 2(5) + 3 = 13$.
• $u_{4} = 2(13) + 3 = 29$.
• $u_{5} = 2(29) + 3 = 61$.

Convergent vs divergent recurrences. A recurrence converges to a limit $L$ when $u_{n + 1} \to L$ and $u_{n} \to L$, so $L = f(L)$. Solving gives the candidate limit.

Example. $u_{n + 1} = \tfrac{1}{2}(u_{n} + 6)$. Limit: $L = \tfrac{1}{2}(L + 6) \Rightarrow 2L = L + 6 \Rightarrow L = 6$. From any starting value the sequence converges to $6$.

Edexcel-style 'show that' questions. Often a recurrence is paired with an explicit formula and you're asked to verify a specific term, or to prove the limit by induction (the induction direction is rare in P2).