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Number & Algebra

Chapter 4 · Topic 1 Exam

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Coverage
Mixed practice over sequences/series, proof, counting & binomial.
Format
18 questions across 4 sections (A: 6, B: 3, C: 5, D: 4 multi-part). Designed for ~60 min in one sitting; pause and resume any time.
Arithmetic — $n$th term
un=u1+(n−1)du_n = u_1 + (n-1)d
Arithmetic — sum to $n$
Sn=n2 ⁣(2u1+(n−1)d)=n2(u1+un)S_n = \tfrac{n}{2}\!\bigl(2u_1 + (n-1)d\bigr) = \tfrac{n}{2}(u_1+u_n)
Two-terms gap (arithmetic)
uq−up=(q−p) du_q - u_p = (q-p)\,d
Use this to recover dd when two non-consecutive terms are given.
Geometric — $n$th term
un=u1 r n−1u_n = u_1\, r^{\,n-1}
Geometric — finite sum
Sn=u1(1−rn)1−r,r≠1S_n = \dfrac{u_1(1 - r^n)}{1 - r},\quad r\ne 1
Geometric — sum to infinity
S∞=u11−r,∣r∣<1  strictlyS_{\infty} = \dfrac{u_1}{1 - r},\quad |r| < 1\;\text{strictly}
Recover term from sum
un=Sn−Sn−1  (n≥2),u1=S1u_n = S_n - S_{n-1}\;(n\ge 2),\quad u_1 = S_1
Sigma notation
∑r=1nar=a1+a2+⋯+an\sum_{r=1}^{n} a_r = a_1 + a_2 + \cdots + a_n
Sigma properties
∑(αar+βbr)=α ⁣∑ar+β ⁣∑br,    ∑r=1nc=nc\sum(\alpha a_r + \beta b_r) = \alpha\!\sum a_r + \beta\!\sum b_r,\;\;\sum_{r=1}^{n} c = nc
Standard sums
∑r=1nr=n(n+1)2,    ∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^{n} r = \tfrac{n(n+1)}{2},\;\;\sum_{r=1}^{n} r^{2} = \tfrac{n(n+1)(2n+1)}{6}
Compound interest
FV=PV (1+i)nFV = PV\,(1 + i)^{n}
ii = period rate (annual rate ÷\div compounds per year). nn = total periods. Depreciation: same formula with 1−i1 - i.
Simple interest
I=PV⋅i⋅nI = PV \cdot i \cdot n
Even / odd algebra
even=2k,    odd=2k+1    (k∈Z)\text{even} = 2k,\;\;\text{odd} = 2k+1\;\;(k\in\mathbb{Z})
Express assumed even/odd integers this way, then manipulate.
Direct proof — sum of two evens
2a+2b=2(a+b)2a + 2b = 2(a+b)
Template: write each as 2k2k, factor 2 out, conclude.
Counter-example — usage
ONE concrete instance is enough to disprove a 'for all' claim. Example: n2+n+41n^{2}+n+41 is prime for n=0,1,…,39n=0,1,\dots,39 but n=40n=40 gives 41241^{2} — not prime.
Contradiction — template
Assume the OPPOSITE of what you want, derive a logical impossibility (e.g. an integer that is both even and odd), conclude the original must hold.
Contradiction — $\sqrt{2}$ irrational
Assume 2=p/q\sqrt{2}=p/q in lowest terms ⇒ p2=2q2p^{2}=2q^{2} ⇒ pp even ⇒ qq even ⇒ contradicts 'lowest terms'.
Induction — structure
P(1)  true;    P(k)⇒P(k+1)  ⇒  P(n)  ∀ n≥1P(1)\;\text{true};\;\;P(k)\Rightarrow P(k+1)\;\Rightarrow\;P(n)\;\forall\,n\ge 1
Base case, inductive hypothesis (IH), inductive step, conclusion. The IH must appear in the step.
Induction — sum example
P(n):∑r=1nr=n(n+1)2P(n): \sum_{r=1}^{n} r = \tfrac{n(n+1)}{2}
Step: add (k+1)(k+1) to IH ⇒ k(k+1)2+(k+1)=(k+1)(k+2)2\tfrac{k(k+1)}{2}+(k+1) = \tfrac{(k+1)(k+2)}{2}.
Method-choice quick guide
'For all n∈Z+n\in\mathbb{Z}^{+}' → induction. 'For all x∈Rx\in\mathbb{R}, PP' → direct. suspected false → counter-example. 'xx irrational' or 'no such xx' → contradiction.
Multiplication principle (AND)
Independent stages with aa, bb outcomes give a×ba \times b total.
Addition principle (OR)
Mutually exclusive alternatives with aa, bb outcomes give a+ba + b total.
Factorial
n!=n(n−1)(n−2)⋯2⋅1,    0!=1n! = n(n-1)(n-2)\cdots 2\cdot 1,\;\; 0! = 1
Arrange $n$ distinct in a row
n!n!
Permutations (order matters)
nPr=n!(n−r)!{}^{n}P_{r} = \dfrac{n!}{(n-r)!}
Combinations (order doesn't)
nCr=(nr)=n!r! (n−r)!{}^{n}C_{r} = \binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}
Identical-objects arrangement
n!n1! n2!⋯nk!\dfrac{n!}{n_{1}!\,n_{2}!\cdots n_{k}!}
Circular arrangement
(n−1)!(n-1)!
Fix one seat to kill rotational symmetry.
Symmetry of $\binom{n}{r}$
(nr)=(nn−r)\binom{n}{r} = \binom{n}{n-r}
Pascal's rule
(nr)=(n−1r−1)+(n−1r)\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}
Binomial theorem
(a+b)n=∑r=0n(nr)an−rbr(a+b)^{n} = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}
General term
Tr+1=(nr) an−r brT_{r+1} = \binom{n}{r}\, a^{n-r}\, b^{r}
Indexed from r=0r=0. Use to pick out a specific term (xkx^{k}, constant term, …).
Coefficient recipe
For (α+βx)n(\alpha + \beta x)^{n}: write Tr+1=(nr)αn−r(βx)rT_{r+1}=\binom{n}{r}\alpha^{n-r}(\beta x)^{r}, set the power of xx equal to the target, solve for rr, evaluate.
Sign care
(a−bx)n  ⇒  Tr+1=(nr)an−r(−b)rxr(a-bx)^{n}\;\Rightarrow\;T_{r+1}=\binom{n}{r}a^{n-r}(-b)^{r}x^{r}
Pick up (−1)r(-1)^{r} in each term.
Trap — geometric infinite
S∞S_{\infty} exists ONLY for ∣r∣<1|r|<1 strict. r=±1r=\pm 1 does NOT converge (Grandi's series).
Trap — compound exponent
If interest is compounded mm times/year for tt years, n=m tn = m\,t and the periodic rate is annualm\tfrac{\text{annual}}{m}.
Trap — binomial power
In (a+bx)n(a+bx)^{n}, each term carries an extra brb^{r} as well as xrx^{r} — don't drop the bb.
Trap — induction
You MUST use the IH inside the step. If the hypothesis P(k)P(k) is never invoked, it's not induction — it's just a direct proof for n=k+1n=k+1.