7.762Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"df3870e7-8d26-496e-ae34-62cdbc6e3932","revisionId":"5a59cd71-6909-4554-8dd8-54f4f63da161","questionNumber":"14","questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":4,"sourceTitle":"Question Set 7: Algebraic Methods","calculatorUse":null,"parts":[{"id":"ce9f79de-6a2d-4df2-9407-49476ed6a461","label":"0","content":"Prove that $(2n + 3)^2 - (2n - 3)^2$ is a multiple of 6 for all values of $n$","markscheme":"- Expand brackets: $(4n^2 + 12n + 9) - (4n^2 - 12n + 9)$. M1\n- Simplify: $24n$. A1\n- Show as a multiple of 6: $6(4n)$. A1\n- Conclude it is a multiple of 6. A1","marks":4,"order":0,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Correctly expands both squared binomials.","code":"M"},{"id":"p2","marks":1,"description":"Simplifies the expression to $24n$.","code":"A"},{"id":"p3","marks":2,"description":"Factors out 6 to prove it is a multiple of 6.","code":"A"}],"finalAnswers":[],"generatedAt":"2026-05-12T02:14:06.612Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"algebraic_proof","label":"Expand and factor","steps":[{"type":"text","order":1,"title":"State the domain","content":"Interpret $n$ as an integer, since “multiple of 6” is an integer divisibility statement. We will prove the result for every integer $n$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=original}{(2n+3)^2-(2n-3)^2}$"},{"latex":"$$\\htmlData{anchor{expanded}}{(4n^2+12n+9)-(4n^2-12n+9)}$"}],"connections":[{"to":"expanded","from":"original","color":"sky","label":"expand"}]},"order":2,"title":"Expand each square","description":"Expand both squares first. This explicitly shows the required expansion working and earns the M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor{expanded}}{(4n^2+12n+9)-(4n^2-12n+9)}$"},{"latex":"$$\\htmlData{anchor{distributed}}{4n^2+12n+9-4n^2+12n-9}$"},{"latex":"$$\\htmlData{anchor{simplified}}{24n}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"expanded"},{"color":"amber","style":"text-color","anchor":"distributed"},{"color":"orange","style":"text-color-with-background","anchor":"simplified"}],"connections":[{"to":"distributed","from":"expanded","color":"sky","label":"distribute the minus","style":"solid"},{"to":"simplified","from":"distributed","color":"orange","label":"collect like terms","style":"solid"}]},"order":3,"title":"Distribute the minus","description":"Distribute the outer minus through the second bracket: every sign changes. In particular, the negative of -12n is +12n, so the two 12n terms add."}]}],"generatedAt":"2026-07-20T16:58:31.859Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"f3b8ac72-eda1-4739-8e55-d08ee77a0240","revisionId":"bf82880e-53b2-425f-bc7c-5ad3af0416c6","questionNumber":"15","questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":4,"sourceTitle":"Question Set 7: Algebraic Methods","calculatorUse":null,"parts":[{"id":"44212239-064f-4645-ba36-4fb0ccda288e","label":"0","content":"Prove that the sum of the squares of two consecutive odd integers is always 2 more than a multiple of 8.","markscheme":"- Define consecutive odd integers: $(2n+1)$ and $(2n+3)$. M1\n- Sum of squares: $(2n+1)^2 + (2n+3)^2 = (4n^2 + 4n + 1) + (4n^2 + 12n + 9)$. M1\n- Simplify: $8n^2 + 16n + 10$. A1\n- Rewrite as $8(n^2 + 2n + 1) + 2$, which is 2 more than a multiple of 8. A1","marks":4,"order":0,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Sets up the sum of squares for two consecutive odd integers.","code":"M"},{"id":"p2","marks":1,"description":"Expands and simplifies the expression to $8n^2 + 16n + 10$.","code":"A"},{"id":"

Algebraic Methods

Question 13

Algebraic Methods Questions

Algebraic Methods

Question 13

Algebraic Methods Questions