Direct Deductive Proof & Algebraic Identities
- Nature of Mathematical Proof: A proof establishes beyond doubt that a statement holds true for all values in its defined domain. Checking numerical examples may inspire a conjecture, but examples never constitute a valid mathematical proof.
- Standard Algebraic Representations:
- Even integer: $2k$, where $k \in \mathbb{Z}$
- Odd integer: $2k + 1$ (or $2k - 1$), where $k \in \mathbb{Z}$
- Consecutive integers: $n, n+1, n+2$
- Consecutive even integers: $2k, 2k+2$; consecutive odd: $2k+1, 2k+3$
- Divisible by $m$: $m \cdot k$, where $k \in \mathbb{Z}$
- Rigorous Direct Proof Protocol: Start strictly with one side (usually the Left-Hand Side, LHS) and use known algebraic manipulations (expansion, factorization, common denominators) until the Right-Hand Side (RHS) is reached. Critical IB Rule: Never assume $\text{LHS} = \text{RHS}$ at the start and manipulate both sides simultaneously—that is circular reasoning and loses marks ($R0$).
Step 1: Define algebraic representations.
Let the two consecutive odd integers be $2k + 1$ and $2k + 3$, where $k \in \mathbb{Z}$.
Step 2: Express the difference of their squares.
$$\text{Difference} = (2k + 3)^2 - (2k + 1)^2$$Step 3: Expand each square algebraically.
$$(2k + 3)^2 = 4k^2 + 12k + 9$$ $$(2k + 1)^2 = 4k^2 + 4k + 1$$Step 4: Subtract and simplify.
$$\text{Difference} = (4k^2 + 12k + 9) - (4k^2 + 4k + 1) = 8k + 8$$Step 5: Factor out the common multiple.
$$\text{Difference} = 8(k + 1)$$Since $k \in \mathbb{Z}$, $k + 1$ is an integer. Thus, $8(k + 1)$ is divisible by 8. Therefore, the difference between the squares of any two consecutive odd integers is always a multiple of 8. $\blacksquare$
Disproof by Counterexample
- Disproving Universal Claims: A conjecture stating that a property holds "for all $x$ in a set $S$" ($\forall x \in S, P(x)$) is disproved by finding a single instance $x_0 \in S$ where the statement is false ($P(x_0)$ is false).
- Requirements for a Valid Counterexample:
- The chosen test value must belong strictly to the stated domain (e.g., if the domain is positive integers, you cannot use $x = -2$ or $x = 0$).
- You must explicitly evaluate both sides or evaluate the expression to clearly show that the claimed result fails.
- Examiner Tip: Counterexamples only disprove universal statements; they cannot prove existence statements or universal laws.
Step 1: Test values within the domain $n \in \mathbb{Z}^+$.
For $n = 1$: $1^2 + 1 + 41 = 43$ (prime).
For $n = 2$: $2^2 + 2 + 41 = 47$ (prime).
Step 2: Identify an algebraic vulnerability where divisibility by 41 is forced.
Notice what happens when $n = 41$:
$$41^2 + 41 + 41 = 41(41 + 1 + 1) = 41 \times 43 = 1763$$Since $n = 41 \in \mathbb{Z}^+$ and $1763$ has factors $1, 41, 43, 1763$, it is composite (not prime).
(Alternatively, $n = 40$ gives $40^2 + 40 + 41 = 40(40 + 1) + 41 = 40(41) + 41 = 41(40 + 1) = 41^2 = 1681$, which is also composite).
Step 3: Conclude clearly.
When $n = 41$, $n^2 + n + 41 = 1763 = 41 \times 43$, which is not prime. A single counterexample disproves the conjecture. $\blacksquare$
Pascal's Triangle & Combinations $\binom{n}{r}$
- Combinations Formula (from Formula Booklet): $$\binom{n}{r} = {}^nC_r = \frac{n!}{r!(n - r)!}, \quad 0 \le r \le n, \; n, r \in \mathbb{N}$$ with $0! = 1$, $\binom{n}{0} = 1$, and $\binom{n}{n} = 1$.
- Symmetry Property: $\binom{n}{r} = \binom{n}{n - r}$. For example, $\binom{7}{2} = \binom{7}{5} = \frac{7 \times 6}{2 \times 1} = 21$.
- Pascal's Addition Identity: $$\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$$ Each interior entry in Pascal's triangle is the sum of the two entries directly above it.
Step 1: Identify coefficients from Row 4 of Pascal's triangle.
For $n = 4$, the coefficients are: $1, 4, 6, 4, 1$.
Step 2: Assign $a = 2x$ and $b = -3$.
$$(2x - 3)^4 = 1(2x)^4 + 4(2x)^3(-3)^1 + 6(2x)^2(-3)^2 + 4(2x)^1(-3)^3 + 1(-3)^4$$Step 3: Evaluate each term carefully (tracking negative signs).
- Term 1: $1 \times 16x^4 = 16x^4$
- Term 2: $4 \times (8x^3) \times (-3) = -96x^3$
- Term 3: $6 \times (4x^2) \times (9) = 216x^2$
- Term 4: $4 \times (2x) \times (-27) = -216x$
- Term 5: $1 \times 81 = 81$
Step 4: Combine into final expanded polynomial.
$$(2x - 3)^4 = 16x^4 - 96x^3 + 216x^2 - 216x + 81$$The Binomial Theorem & The General Term Formula
- The Binomial Theorem (Formula Booklet): $$(a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \dots + \binom{n}{r}a^{n-r}b^r + \dots + b^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r}b^r$$ for $n \in \mathbb{N}$. Total number of terms in $(a+b)^n$ is $n + 1$.
- The General Term ($T_{r+1}$): $$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$ Index offset: The 1st term has $r = 0$, the 2nd term has $r = 1$, the $(k)$th term has $r = k - 1$.
- Strategy for Specific Powers & Terms Independent of $x$:
- Substitute the specific expressions for $a$ and $b$ into $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
- Separate numerical coefficients from variables to gather all powers of $x$: $x^{p(n-r)} \cdot x^{q r} = x^{p(n-r) + q r}$.
- Set the total exponent of $x$ equal to the requested target power (e.g., set exponent $= 0$ for "term independent of $x$").
- Solve for integer $r$ ($0 \le r \le n$), then evaluate the full term.
Step 1: Set up the general term formula.
Here $n = 6$, $a = 2x^2$, and $b = -3x^{-1}$.
$$T_{r+1} = \binom{6}{r} (2x^2)^{6-r} \left(-\frac{3}{x}\right)^r$$Step 2: Separate constants and collect powers of $x$.
$$T_{r+1} = \binom{6}{r} 2^{6-r} (-3)^r \cdot (x^2)^{6-r} (x^{-1})^r$$ $$T_{r+1} = \binom{6}{r} 2^{6-r} (-3)^r \cdot x^{2(6-r) - r} = \binom{6}{r} 2^{6-r} (-3)^r \cdot x^{12 - 3r}$$Step 3: Solve for $r$ for the term independent of $x$ (exponent $= 0$).
$$12 - 3r = 0 \implies 3r = 12 \implies r = 4$$Step 4: Substitute $r = 4$ to calculate the term.
$$T_5 = \binom{6}{4} 2^{6-4} (-3)^4 \cdot x^0$$ $$\binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$$ $$T_5 = 15 \times 2^2 \times 81 = 15 \times 4 \times 81 = 60 \times 81 = 4860$$Answer: The term independent of $x$ is $4860$.
Products of Binomials & Solving for Unknown Constants
- Composite Expansions: $(c + dx)(a + bx)^n$: To find the coefficient of $x^k$ in such a product, expand only the necessary terms of $(a + bx)^n$: $$(c + dx) \left[ \dots + A x^{k-1} + B x^k + \dots \right]$$ The term containing $x^k$ arises from two distinct pairings: $$\text{Term in } x^k = (c \cdot B x^k) + (dx \cdot A x^{k-1}) = (cB + dA) x^k$$
- Working Backwards to Find an Unknown Parameter: When an exam question states that "the coefficient of $x^k$ is $C$", use the general term to form an equation in terms of the unknown parameter $a$ or $k$, then solve algebraically.
Step 1: Expand $(2 - x)^5$ up to the $x^2$ term.
$$(2 - x)^5 = \binom{5}{0} 2^5 + \binom{5}{1} 2^4 (-x)^1 + \binom{5}{2} 2^3 (-x)^2 + \dots$$ $$(2 - x)^5 = 1(32) + 5(16)(-x) + 10(8)(x^2) + \dots$$ $$(2 - x)^5 = 32 - 80x + 80x^2 + \dots$$Step 2: Multiply by $(1 + kx)$ and extract terms that yield $x^2$.
$$(1 + kx)(32 - 80x + 80x^2 + \dots)$$ $$\text{Terms in } x^2: \quad 1 \cdot (80x^2) + kx \cdot (-80x) = (80 - 80k)x^2$$Step 3: Set equal to the given coefficient and solve for $k$.
$$80 - 80k = 40 \implies 80k = 40 \implies k = \frac{40}{80} = \frac{1}{2}$$Answer: $k = \frac{1}{2}$.
10 Progressive Practice Exercises
Tier 1 tests direct algebraic proof and standard expansions; Tier 2 introduces exam-style general terms and counterexamples; Tier 3 challenges composite products and multi-step proofs.
Tier 1 Core Fluency (Problems 1–3)
Prove algebraically that the sum of any three consecutive integers is always divisible by 3.
View Full Worked Solution
Step 1: Let the three consecutive integers be $n, n+1$, and $n+2$, where $n \in \mathbb{Z}$.
Step 2: Write an expression for their sum:
$$\text{Sum} = n + (n + 1) + (n + 2) = 3n + 3$$Step 3: Factor out 3:
$$\text{Sum} = 3(n + 1)$$Since $n$ is an integer, $n + 1$ is an integer. Thus, $3(n + 1)$ is a multiple of 3. Therefore, the sum of any three consecutive integers is always divisible by 3. $\blacksquare$
Expand $\left(x + \frac{2}{x}\right)^4$ completely and simplify each term.
View Full Worked Solution
Step 1: Use binomial coefficients for $n = 4$: $1, 4, 6, 4, 1$.
$$\left(x + \frac{2}{x}\right)^4 = 1(x)^4 + 4(x)^3\left(\frac{2}{x}\right)^1 + 6(x)^2\left(\frac{2}{x}\right)^2 + 4(x)^1\left(\frac{2}{x}\right)^3 + 1\left(\frac{2}{x}\right)^4$$Step 2: Simplify each term:
- Term 1: $x^4$
- Term 2: $4 x^3 \cdot \frac{2}{x} = 8x^2$
- Term 3: $6 x^2 \cdot \frac{4}{x^2} = 24$
- Term 4: $4 x \cdot \frac{8}{x^3} = \frac{32}{x^2}$
- Term 5: $\frac{16}{x^4}$
Answer: $x^4 + 8x^2 + 24 + \frac{32}{x^2} + \frac{16}{x^4}$.
Find the coefficient of $x^3$ in the binomial expansion of $(3 - 2x)^5$.
View Full Worked Solution
Step 1: The general term is $T_{r+1} = \binom{5}{r} (3)^{5-r} (-2x)^r = \binom{5}{r} 3^{5-r} (-2)^r x^r$.
Step 2: For the coefficient of $x^3$, set $r = 3$:
$$T_4 = \binom{5}{3} 3^{5-3} (-2)^3 x^3 = \binom{5}{3} 3^2 (-8) x^3$$Step 3: Calculate $\binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10$.
$$T_4 = 10 \times 9 \times (-8) x^3 = -720 x^3$$Answer: The coefficient of $x^3$ is $-720$.
Tier 2 Exam Modeling (Problems 4–7)
Consider the statement: "If $n$ is an integer and $n^2$ is divisible by 4, then $n$ must be divisible by 4." Disprove this statement by giving a counterexample.
View Full Worked Solution
Step 1: We need an integer $n$ such that $n^2$ is divisible by 4, but $n$ is NOT divisible by 4.
Step 2: Test $n = 2$ ($2 \in \mathbb{Z}$):
$$n^2 = 2^2 = 4$$$4$ is divisible by 4 ($4 = 4 \times 1$).
However, $n = 2$ is clearly NOT divisible by 4 ($2 / 4 = 0.5 \notin \mathbb{Z}$).
Answer: Let $n = 2$ (or $n = 6, 10, \dots$). Then $n^2 = 4$, which is divisible by 4, but $n = 2$ is not divisible by 4. This counterexample disproves the statement. $\blacksquare$
Find the term independent of $x$ in the binomial expansion of $\left(3x - \frac{2}{x^2}\right)^6$.
View Full Worked Solution
Step 1: Write the general term:
$$T_{r+1} = \binom{6}{r} (3x)^{6-r} \left(-\frac{2}{x^2}\right)^r = \binom{6}{r} 3^{6-r} (-2)^r \cdot x^{6-r} (x^{-2})^r$$ $$T_{r+1} = \binom{6}{r} 3^{6-r} (-2)^r \cdot x^{6 - 3r}$$Step 2: Set the exponent of $x$ to $0$:
$$6 - 3r = 0 \implies 3r = 6 \implies r = 2$$Step 3: Substitute $r = 2$:
$$T_3 = \binom{6}{2} 3^{6-2} (-2)^2 = 15 \times 3^4 \times 4 = 15 \times 81 \times 4 = 60 \times 81 = 4860$$Answer: The term independent of $x$ is $4860$.
In the expansion of $(2 + ax)^7$, where $a > 0$, the coefficient of $x^3$ is $2240$. Find the value of the constant $a$.
View Full Worked Solution
Step 1: Find the general term containing $x^3$ ($r = 3$):
$$T_4 = \binom{7}{3} 2^{7-3} (ax)^3 = \binom{7}{3} 2^4 a^3 x^3$$Step 2: Evaluate the numerical components:
$$\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$ $$2^4 = 16$$ $$\text{Coefficient of } x^3 = 35 \times 16 \times a^3 = 560 a^3$$Step 3: Equate to the given coefficient and solve for $a$:
$$560 a^3 = 2240 \implies a^3 = \frac{2240}{560} = 4 \implies a = \sqrt[3]{4}$$Answer: $a = \sqrt[3]{4}$.
Prove that $(2n + 1)^2 - 1$ is always a multiple of 8 for any positive integer $n$.
View Full Worked Solution
Step 1: Expand the algebraic expression:
$$(2n + 1)^2 - 1 = (4n^2 + 4n + 1) - 1 = 4n^2 + 4n$$Step 2: Factor the expression:
$$4n^2 + 4n = 4n(n + 1)$$Step 3: Analyze the product $n(n + 1)$:
Notice that $n$ and $n + 1$ are two consecutive integers. In any pair of consecutive integers, exactly one of them is even. Therefore, $n(n + 1)$ is always divisible by 2. That is, $n(n + 1) = 2k$ for some integer $k$.
Step 4: Substitute back:
$$4n(n + 1) = 4(2k) = 8k$$Since $k \in \mathbb{Z}$, $8k$ is a multiple of 8. Therefore, $(2n + 1)^2 - 1$ is always a multiple of 8 for any positive integer $n$. $\blacksquare$
Tier 3 Complex Applications & Synthesis (Problems 8–10)
In the expansion of $(2x + 1)(x - 3)^6$, find the coefficient of $x^4$.
View Full Worked Solution
Step 1: Write the general term of $(x - 3)^6$:
$$T_{r+1} = \binom{6}{r} x^{6-r} (-3)^r$$To obtain an $x^4$ term when multiplied by $(2x + 1)$: $$(2x + 1) \left[ \dots + A x^3 + B x^4 + \dots \right]$$ The $x^4$ term comes from: $$\text{Term in } x^4 = 2x \cdot (A x^3) + 1 \cdot (B x^4) = (2A + B) x^4$$
Step 2: Find $A$ (the coefficient of $x^3$ in $(x-3)^6$): For $x^3$, $6 - r = 3 \implies r = 3$: $$A = \binom{6}{3} (-3)^3 = 20 \times (-27) = -540$$
Step 3: Find $B$ (the coefficient of $x^4$ in $(x-3)^6$): For $x^4$, $6 - r = 4 \implies r = 2$: $$B = \binom{6}{2} (-3)^2 = 15 \times 9 = 135$$
Step 4: Combine coefficients: $$\text{Coefficient of } x^4 = 2A + B = 2(-540) + 135 = -1080 + 135 = -945$$
Answer: The coefficient of $x^4$ is $-945$.
Given that the coefficient of $x^2$ in the expansion of $\left(kx^2 + \frac{1}{x}\right)^7$ is equal to $280$, find the exact value of the constant $k$.
View Full Worked Solution
Step 1: Write the general term:
$$T_{r+1} = \binom{7}{r} (kx^2)^{7-r} \left(\frac{1}{x}\right)^r = \binom{7}{r} k^{7-r} \cdot x^{2(7-r) - r} = \binom{7}{r} k^{7-r} \cdot x^{14 - 3r}$$Step 2: Set the exponent equal to 2:
$$14 - 3r = 2 \implies 3r = 12 \implies r = 4$$Step 3: Set up the coefficient equation with $r = 4$:
$$\text{Coefficient} = \binom{7}{4} k^{7-4} = \binom{7}{3} k^3 = 35 k^3$$Step 4: Equate to 280 and solve for $k$:
$$35 k^3 = 280 \implies k^3 = \frac{280}{35} = 8 \implies k = \sqrt[3]{8} = 2$$Answer: $k = 2$.
(a) Prove the algebraic identity $(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2$ for all real numbers $a, b, c, d$.
(b) Hence, by expressing $65 = (2^2 + 1^2)(3^2 + 2^2)$, write $65$ as the sum of two positive square numbers in two distinct ways.
View Full Worked Solution
Part (a): Direct Proof of Identity
Start with the Right-Hand Side (RHS):
$$\text{RHS} = (ac - bd)^2 + (ad + bc)^2$$ $$\text{RHS} = (a^2c^2 - 2abcd + b^2d^2) + (a^2d^2 + 2abcd + b^2c^2)$$Notice that $-2abcd$ and $+2abcd$ cancel out:
$$\text{RHS} = a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2$$Group terms by factoring:
$$\text{RHS} = a^2(c^2 + d^2) + b^2(c^2 + d^2) = (a^2 + b^2)(c^2 + d^2) = \text{LHS}$$Since $\text{RHS} = \text{LHS}$, the identity is proved. $\blacksquare$
Part (b): Application to 65
Let $a = 2, b = 1, c = 3, d = 2$:
$$65 = (2^2 + 1^2)(3^2 + 2^2)$$Using the identity:
$$65 = (ac - bd)^2 + (ad + bc)^2 = (2(3) - 1(2))^2 + (2(2) + 1(3))^2 = (6 - 2)^2 + (4 + 3)^2 = 4^2 + 7^2 = 16 + 49 = 65$$Alternatively, interchange $c$ and $d$ (let $c = 2, d = 3$):
$$65 = (2(2) - 1(3))^2 + (2(3) + 1(2))^2 = (4 - 3)^2 + (6 + 2)^2 = 1^2 + 8^2 = 1 + 64 = 65$$Answer: $65 = 4^2 + 7^2$ and $65 = 1^2 + 8^2$.
Binomial Expansion & Combinations
- Binomial Expansion: $$(a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \dots + \binom{n}{r}a^{n-r}b^r + \dots + b^n$$ Where $n \in \mathbb{N}$ and $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.
- General Term: $$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$ Note that the $(r+1)$th term has power $b^r$.
Proof Definitions & Combinatorial Identities
- Standard Proof Formats: $$\text{Even: } 2k, \quad \text{Odd: } 2k+1, \quad \text{Consecutive: } n, n+1$$ Always declare domain: "where $k \in \mathbb{Z}$".
- Pascal's Identity & Symmetry: $$\binom{n}{r} = \binom{n}{n-r}, \qquad \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$$
- Total Number of Terms in Expansion: $$\text{Number of terms in } (a + b)^n = n + 1$$
- Sum of Binomial Coefficients: $$\sum_{r=0}^n \binom{n}{r} = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n \quad (\text{substitute } a=1, b=1)$$
Critical IB Examination Pitfalls
- Circular Reasoning in Direct Proofs: Writing $\text{LHS} = \text{RHS}$ at line 1 and simplifying both sides simultaneously is circular reasoning. In IB markschemes, you must start from the LHS and demonstrate that it identically reduces to the RHS (or vice versa).
- Index Offset Error ($r$ vs Term Number): The 5th term is $T_5$, which corresponds to $r = 4$, NOT $r = 5$. A very common mistake is using $r = 5$ for the 5th term.
- Losing Negative Signs in Expansions: In $(2x - 3y)^5$, the second component is $b = (-3y)$. When $r$ is odd, $(-3y)^r$ is negative. Always enclose negative terms in parentheses: $(-3y)^r = (-3)^r y^r$.
- Forgetting the Coefficient Power: When expanding $(3x^2)^4$, write $3^4 (x^2)^4 = 81x^8$. Students often forget to apply the power to the numerical coefficient, writing $3x^8$.
- Invalid Counterexamples Outside the Domain: If a conjecture states "for all positive integers $n$", testing $n = 0$ or $n = -1$ earns zero marks because those values lie outside the specified domain.
IB AA SL Topic Check: Simple Proof & Binomial Theorem
Section A tests instant diagnostic recognition with automated feedback; Section B assesses Paper 1 and Paper 2 style multi-mark exam problems with full markscheme breakdown.
Section A: Diagnostic Concept Check
Score: 0 / 4Question 1: Which of the following values of $n$ serves as a valid counterexample to disprove the conjecture: "For all positive integers $n$, $2^n + 1$ is a prime number"?
Question 2: In the expansion of $\left(x^3 - \frac{2}{x}\right)^8$, what value of $r$ yields the term independent of $x$?
Question 3: What is the coefficient of $x^2$ in the expansion of $(1 - 3x)^4$?
Question 4: What is the algebraic value of $(n+1)^2 - (n-1)^2$ for any integer $n$, proving it is always a multiple of 4?
Section B: Exam-Style Free Response Questions
Question 5: Product with Unknown Parameter
Consider the expansion of $(1 + ax)(2 - x)^6$, where $a$ is a non-zero constant.
(a) Find the first three terms, in ascending powers of $x$, of the expansion of $(2 - x)^6$. [3 marks]
(b) Given that the coefficient of $x^2$ in the expansion of $(1 + ax)(2 - x)^6$ is $120$, find the value of $a$. [3 marks]
View Official Markscheme & Model Solution
Part (a) [3 Marks]:
$$(2 - x)^6 = \binom{6}{0}2^6 + \binom{6}{1}2^5(-x) + \binom{6}{2}2^4(-x)^2 + \dots$$- $\binom{6}{0} 2^6 = 1 \times 64 = 64$ (A1)
- $\binom{6}{1} 2^5 (-x) = 6 \times 32 \times (-x) = -192x$ (A1)
- $\binom{6}{2} 2^4 (-x)^2 = 15 \times 16 \times x^2 = 240x^2$ (A1)
So $(2 - x)^6 = 64 - 192x + 240x^2 + \dots$
Part (b) [3 Marks]:
$$(1 + ax)(64 - 192x + 240x^2 + \dots)$$The terms in $x^2$ are:
$$1 \cdot (240x^2) + ax \cdot (-192x) = (240 - 192a)x^2 \quad \text{\bf (M1)(A1)}$$Equating to 120:
$$240 - 192a = 120 \implies 192a = 120 \implies a = \frac{120}{192} = \frac{5}{8} \quad \text{\bf (A1)}$$Answer: $a = \frac{5}{8}$ (or $0.625$).
Question 6: Proof & Independent Term
(a) Prove algebraically that the sum of the squares of any two consecutive integers is always an odd number. [3 marks]
(b) In the expansion of $\left(x^2 + \frac{k}{x}\right)^9$, where $k > 0$, the term independent of $x$ is $672$. Find the exact value of $k$. [4 marks]
View Official Markscheme & Model Solution
Part (a) [3 Marks]:
Let the two consecutive integers be $n$ and $n + 1$, where $n \in \mathbb{Z}$. (M1)
$$\text{Sum of squares} = n^2 + (n + 1)^2 = n^2 + (n^2 + 2n + 1) = 2n^2 + 2n + 1 \quad \text{\bf (A1)}$$ $$\text{Sum of squares} = 2(n^2 + n) + 1$$Since $n \in \mathbb{Z}$, $n^2 + n$ is an integer (let $m = n^2 + n$).
Then the sum is $2m + 1$, which by definition is odd for all $m \in \mathbb{Z}$. (R1) $\blacksquare$
Part (b) [4 Marks]:
The general term is:
$$T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(\frac{k}{x}\right)^r = \binom{9}{r} k^r x^{18 - 2r - r} = \binom{9}{r} k^r x^{18 - 3r} \quad \text{\bf (M1)(A1)}$$For the term independent of $x$, set exponent $= 0$:
$$18 - 3r = 0 \implies r = 6 \quad \text{\bf (A1)}$$Substitute $r = 6$ into the term:
$$T_7 = \binom{9}{6} k^6 = \binom{9}{3} k^6 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} k^6 = 84 k^6$$Set equal to 672 and solve for $k$:
$$84 k^6 = 672 \implies k^6 = \frac{672}{84} = 8 \quad \text{\bf (M1)}$$ $$k = 8^{1/6} = (2^3)^{1/6} = 2^{3/6} = 2^{1/2} = \sqrt{2} \quad \text{\bf (A1)}$$Answer: $k = \sqrt{2}$.
Topic 1 Curriculum
All Number and Algebra Study Guides
About this guide
Independent SJMaths resource; not an official IB publication. Follows the AA syllabus first assessed in 2021 (final assessment: November 2028). Use the separate 2029 guide for the new course.
Check the IB Mathematics programme page and the official AA subject guide for the authoritative syllabus and examination-session requirements.