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$w$, which is given by $(-3)^6$ or $\\binom{6}{6}(-3)^6$ M1\n- Evaluates the constant term to be $729$ A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substituting w = 0","steps":[{"type":"text","order":0,"title":"Understand constant terms","content":"In any polynomial expression, the constant term is the term that does not depend on the variable.\n\nIf we expand $R(w)$ binomially, we get a series of terms of the form $a_k w^k$:\n\n$$R(w) = a_6 w^6 + a_5 w^5 + \\dots + a_1 w + a_0$$\n\nHere, $a_0$ is the constant term. Notice that if we substitute $w = 0$ into the expanded function, all terms with a power of $w$ disappear, leaving only the constant term:\n\n$$R(0) = a_0$$","highlights":[{"text":"constant term","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$R(\\htmlData{anchor=w-val}{\\color{@sky}{0}}) = \\left( \\frac{1}{2}(\\htmlData{anchor=w-sub}{\\color{@sky}{0}}) - 3 \\right)^6$"},{"latex":"$$\\phantom{R(0)} = (\\htmlData{anchor=result}{\\color{@amber}{-3}})^6$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"w-val"},{"color":"amber","style":"box","anchor":"result"}],"connections":[{"to":"w-sub","from":"w-val","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Substitute w = 0","description":"We can find the constant term directly by substituting $w = 0$ into our original function, $R(w) = \\left( \\frac{1}{2}w - 3 \\right)^6$."},{"type":"text","order":2,"title":"Evaluate the power","content":"Now we compute the value of $(-3)^6$:\n\n$$(-3)^6 = 729$$\n\nSince the exponent $6$ is even, the result is positive.\n\nEvaluating this correctly earns the final accuracy mark."}]},{"id":"binomial_theorem","label":"Using the Binomial Theorem","steps":[{"type":"text","order":0,"title":"State the general term","content":"To find the constant term in the binomial expansion of $R(w) = \\left( \\frac{1}{2}w - 3 \\right)^6$, we start by using the general term formula from the Binomial Theorem.\n\nFor an expansion of $(a + b)^n$, the general term is given by:\n\n$$T_{r+1} = \\binom{n}{r} a^{n-r} b^r$$\n\nIn our function, we have:\n- $a = \\frac{1}{2}w$\n- $b = -3$\n- $n = 6$\n\nSubstituting these into our general term formula gives:\n\n$$T_{r+1} = \\binom{6}{r} \\left(\\frac{1}{2}w\\right)^{6-r} (-3)^r$$"},{"type":"text","order":1,"title":"Identify the constant term","content":"The constant term is the term that is independent of $w$, which means the power of $w$ must be equal to zero.\n\nLet's isolate the variable $w$ in our general term expression:\n\n$$T_{r+1} = \\binom{6}{r} \\left(\\frac{1}{2}\\right)^{6-r} w^{6-r} (-3)^r$$\n\nFor the power of $w$ to be zero, we set:\n\n$$6 - r = 0 \\implies r = 6$$\n\nThis tells us that the constant term occurs when $r = 6$, which corresponds to the final term of the expansion."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$T_{7} = \\binom{6}{\\htmlData{anchor=r-val}{\\color{@sky}{6}}} \\left(\\frac{1}{2}w\\right)^{6-\\htmlData{anchor=r-sub1}{\\color{@sky}{6}}} (\\htmlData{anchor=base}{\\color{@amber}{-3}})^{\\htmlData{anchor=r-sub2}{\\color{@sky}{6}}}$"},{"latex":"$$T_{7} = 1 \\cdot \\left(\\frac{1}{2}w\\right)^0 \\cdot \\color{@amber}{(-3)}^{\\color{@sky}{6}}$"},{"latex":"$$T_{7} = 1 \\cdot 1 \\cdot \\htmlData{anchor=final-ans}{\\color{@purple}{729}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"r-val"},{"color":"amber","style":"text-color","anchor":"base"},{"color":"purple","style":"box","anchor":"final-ans"}],"connections":[{"to":"r-sub1","from":"r-val","color":"sky","label":null,"style":"solid"},{"to":"r-sub2","from":"r-val","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Evaluate the constant term","description":"Substitute $r = 6$ back into our general term expression and compute the numerical value to get the constant term."},{"type":"text","order":3,"title":"Confirm the final answer","content":"The constant term in the expansion is **$729$**.\n\n**Markscheme Guidance:**\n- **1 mark (M1)** is awarded for showing the method of recognizing that the constant term is independent of $w$, written as $(-3)^6$ or $\\binom{6}{6}(-3)^6$.\n- **1 mark (A1)** is awarded for correctly evaluating the constant term to be $729$."}]}],"generatedAt":"2026-06-09T09:02:13.768Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"f8c100a1-5033-4d6a-b1fa-6107fc19c0f5:substitution","label":"Substituting w = 0","steps":[{"type":"text","order":0,"title":"Understand constant terms","content":"In any polynomial expression, the constant term is the term that does not depend on the variable.\n\nIf we expand $R(w)$ binomially, we get a series of terms of the form $a_k w^k$:\n\n$$R(w) = a_6 w^6 + a_5 w^5 + \\dots + a_1 w + a_0$$\n\nHere, $a_0$ is the constant term. Notice that if we substitute $w = 0$ into the expanded function, all terms with a power of $w$ disappear, leaving only the constant term:\n\n$$R(0) = a_0$$","highlights":[{"text":"constant term","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$R(\\htmlData{anchor=w-val}{\\color{@sky}{0}}) = \\left( \\frac{1}{2}(\\htmlData{anchor=w-sub}{\\color{@sky}{0}}) - 3 \\right)^6$"},{"latex":"$$\\phantom{R(0)} = (\\htmlData{anchor=result}{\\color{@amber}{-3}})^6$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"w-val"},{"color":"amber","style":"box","anchor":"result"}],"connections":[{"to":"w-sub","from":"w-val","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Substitute w = 0","description":"We can find the constant term directly by substituting $w = 0$ into our original function, $R(w) = \\left( \\frac{1}{2}w - 3 \\right)^6$."},{"type":"text","order":2,"title":"Evaluate the power","content":"Now we compute the value of $(-3)^6$:\n\n$$(-3)^6 = 729$$\n\nSince the exponent $6$ is even, the result is positive.\n\nEvaluating this correctly earns the final accuracy mark."}]},{"id":"f8c100a1-5033-4d6a-b1fa-6107fc19c0f5:binomial_theorem","label":"Using the Binomial Theorem","steps":[{"type":"text","order":0,"title":"State the general term","content":"To find the constant term in the binomial expansion of $R(w) = \\left( \\frac{1}{2}w - 3 \\right)^6$, we start by using the general term formula from the Binomial Theorem.\n\nFor an expansion of $(a + b)^n$, the general term is given by:\n\n$$T_{r+1} = \\binom{n}{r} a^{n-r} b^r$$\n\nIn our function, we have:\n- $a = \\frac{1}{2}w$\n- $b = -3$\n- $n = 6$\n\nSubstituting these into our general term formula gives:\n\n$$T_{r+1} = \\binom{6}{r} \\left(\\frac{1}{2}w\\right)^{6-r} (-3)^r$$"},{"type":"text","order":1,"title":"Identify the constant term","content":"The constant term is the term that is independent of $w$, which means the power of $w$ must be equal to zero.\n\nLet's isolate the variable $w$ in our general term expression:\n\n$$T_{r+1} = \\binom{6}{r} \\left(\\frac{1}{2}\\right)^{6-r} w^{6-r} (-3)^r$$\n\nFor the power of $w$ to be zero, we set:\n\n$$6 - r = 0 \\implies r = 6$$\n\nThis tells us that the constant term occurs when $r = 6$, which corresponds to the final term of the expansion."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$T_{7} = \\binom{6}{\\htmlData{anchor=r-val}{\\color{@sky}{6}}} \\left(\\frac{1}{2}w\\right)^{6-\\htmlData{anchor=r-sub1}{\\color{@sky}{6}}} (\\htmlData{anchor=base}{\\color{@amber}{-3}})^{\\htmlData{anchor=r-sub2}{\\color{@sky}{6}}}$"},{"latex":"$$T_{7} = 1 \\cdot \\left(\\frac{1}{2}w\\right)^0 \\cdot \\color{@amber}{(-3)}^{\\color{@sky}{6}}$"},{"latex":"$$T_{7} = 1 \\cdot 1 \\cdot \\htmlData{anchor=final-ans}{\\color{@purple}{729}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"r-val"},{"color":"amber","style":"text-color","anchor":"base"},{"color":"purple","style":"box","anchor":"final-ans"}],"connections":[{"to":"r-sub1","from":"r-val","color":"sky","label":null,"style":"solid"},{"to":"r-sub2","from":"r-val","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Evaluate the constant term","description":"Substitute $r = 6$ back into our general term expression and compute the numerical value to get the constant term."},{"type":"text","order":3,"title":"Confirm the final answer","content":"The constant term in the expansion is **$729$**.\n\n**Markscheme Guidance:**\n- **1 mark (M1)** is awarded for showing the method of recognizing that the constant term is independent of $w$, written as $(-3)^6$ or $\\binom{6}{6}(-3)^6$.\n- **1 mark (A1)** is awarded for correctly evaluating the constant term to be $729$."}]}],"generatedAt":"2026-06-09T09:02:23.326Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"f997d6e3-3ccb-42e3-b56b-643b47a8d805","revisionId":"24f29b63-b9e2-4429-9633-5a272b0091c5","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"easy","tier":null,"totalMarks":2,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"80ce58a7-60b2-465c-8ad8-6d7f15d3a0c6","label":null,"content":"State the range of values of \$x\$ for which the binomial expansion of\n\\[\\sqrt[3]{1 - \\frac{x}{5}}\\]\nis valid.","markscheme":"- Recognises that for the expansion of $(1 + u)^n$ to be valid, $|u| < 1$ M1\n- Sets up the inequality $\\left| -\\frac{x}{5} \\right| < 1$ and deduces $|x| < 5$ A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"absolute_value","label":"Absolute Value Inequality","steps":[{"type":"text","order":0,"title":"Identify the validity condition","content":"To find the range of values for which the binomial expansion is valid, we use the standard convergence condition. For an expansion of the form $(1 + u)^n$ where $n$ is not a positive integer, the series converges and is valid if and only if $|u| < 1$."},{"type":"text","order":1,"title":"Set up the inequality","content":"In our given expression, $\\sqrt[3]{1 - \\frac{x}{5}} = \\left(1 - \\frac{x}{5}\\right)^{\\frac{1}{3}}$, the variable term is $u = -\\frac{x}{5}$. Substituting this into our convergence condition gives the absolute value inequality: $\\left|-\\frac{x}{5}\\right| < 1$."},{"type":"text","order":2,"title":"Solve the inequality","content":"We can simplify this absolute value inequality. Since $\\left|-\\frac{x}{5}\\right| = \\frac{|x|}{5}$, our inequality becomes $\\frac{|x|}{5} < 1$. Multiplying both sides by $5$, we get $|x| < 5$. This can also be written as $-5 < x < 5$."}]},{"id":"compound_inequality","label":"Compound Inequality","steps":[{"type":"text","order":0,"title":"Identify the validity condition","content":"For a binomial expansion of the form $(1 + u)^n$, where $n$ is a fraction or negative integer, the expansion is valid if and only if:\n\n$$-1 < u < 1$$\n\nIn the expression $\\sqrt[3]{1 - \\frac{x}{5}} = \\left(1 - \\frac{x}{5}\\right)^{\\frac{1}{3}}$, our term $u$ is:\n\n$$u = -\\frac{x}{5}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$-1 < \\htmlData{anchor=u-term}{\\color{@sky}{u}} < 1$"},{"latex":"$$-1 < \\htmlData{anchor=sub-term}{\\color{@sky}{-\\frac{x}{5}}} < 1$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"sky","style":"text-color","anchor":"sub-term"}],"connections":[{"to":"sub-term","from":"u-term","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Set up compound inequality","description":"We replace $u$ in our compound inequality with the term $-\\frac{x}{5}$ to find the range of validity."},{"type":"text","order":2,"title":"Solve for x","content":"To solve for $x$, we multiply all parts of the compound inequality by $-5$.\n\nRemember, when multiplying or dividing an inequality by a negative number, we **must reverse** the direction of the inequality signs:\n\n$$(-1) \\times (-5) > -\\frac{x}{5} \\times (-5) > 1 \\times (-5)$$\n\nSimplifying this gives:\n\n$$5 > x > -5$$"},{"type":"text","order":3,"title":"State final range","content":"We can rewrite the inequality $5 > x > -5$ in standard ascending order:\n\n$$-5 < x < 5$$\n\nAlternatively, this can be written using modulus notation as:\n\n$$|x| < 5$$"}]}],"generatedAt":"2026-06-09T10:24:21.468Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"80ce58a7-60b2-465c-8ad8-6d7f15d3a0c6:absolute_value","label":"Absolute Value Inequality","steps":[{"type":"text","order":0,"title":"Identify the validity condition","content":"To find the range of values for which the binomial expansion is valid, we use the standard convergence condition. For an expansion of the form $(1 + u)^n$ where $n$ is not a positive integer, the series converges and is valid if and only if $|u| < 1$."},{"type":"text","order":1,"title":"Set up the inequality","content":"In our given expression, $\\sqrt[3]{1 - \\frac{x}{5}} = \\left(1 - \\frac{x}{5}\\right)^{\\frac{1}{3}}$, the variable term is $u = -\\frac{x}{5}$. Substituting this into our convergence condition gives the absolute value inequality: $\\left|-\\frac{x}{5}\\right| < 1$."},{"type":"text","order":2,"title":"Solve the inequality","content":"We can simplify this absolute value inequality. Since $\\left|-\\frac{x}{5}\\right| = \\frac{|x|}{5}$, our inequality becomes $\\frac{|x|}{5} < 1$. Multiplying both sides by $5$, we get $|x| < 5$. This can also be written as $-5 < x < 5$."}]},{"id":"80ce58a7-60b2-465c-8ad8-6d7f15d3a0c6:compound_inequality","label":"Compound Inequality","steps":[{"type":"text","order":0,"title":"Identify the validity condition","content":"For a binomial expansion of the form $(1 + u)^n$, where $n$ is a fraction or negative integer, the expansion is valid if and only if:\n\n$$-1 < u < 1$$\n\nIn the expression $\\sqrt[3]{1 - \\frac{x}{5}} = \\left(1 - \\frac{x}{5}\\right)^{\\frac{1}{3}}$, our term $u$ is:\n\n$$u = -\\frac{x}{5}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$-1 < \\htmlData{anchor=u-term}{\\color{@sky}{u}} < 1$"},{"latex":"$$-1 < \\htmlData{anchor=sub-term}{\\color{@sky}{-\\frac{x}{5}}} < 1$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"sky","style":"text-color","anchor":"sub-term"}],"connections":[{"to":"sub-term","from":"u-term","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Set up compound inequality","description":"We replace $u$ in our compound inequality with the term $-\\frac{x}{5}$ to find the range of validity."},{"type":"text","order":2,"title":"Solve for x","content":"To solve for $x$, we multiply all parts of the compound inequality by $-5$.\n\nRemember, when multiplying or dividing an inequality by a negative number, we **must reverse** the direction of the inequality signs:\n\n$$(-1) \\times (-5) > -\\frac{x}{5} \\times (-5) > 1 \\times (-5)$$\n\nSimplifying this gives:\n\n$$5 > x > -5$$"},{"type":"text","order":3,"title":"State final range","content":"We can rewrite the inequality $5 > x > -5$ in standard ascending order:\n\n$$-5 < x < 5$$\n\nAlternatively, this can be written using modulus notation as:\n\n$$|x| < 5$$"}]}],"generatedAt":"2026-06-09T10:24:32.595Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__ccea","summaries":[{"id":"0429ea80-c16f-442c-a8c7-9db5d90fae0d","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"3950bc86-f600-4c92-a793-037add492875","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"567a339e-ed5f-49fa-b1f0-ea25bac54256","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"5d65fca2-3aa5-4015-9b4d-9e3550438b7c","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"a040b0b7-7843-47a7-b7ee-725bb314c8c1","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"d2616403-974a-4433-8a7d-26f3adfdedb8","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"efd5539a-7834-498d-90fa-afdf1d655594","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"fddc2bfd-83f8-4ddb-8167-514e0cb2afdd","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"d4d32f37-b31e-4754-8a1e-d4ec4eb9e593","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"039ad70c-5bfd-43d0-9b1e-b3edcc8e6a18","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"584871c6-110d-429e-8e67-ba48f1b8a028","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"6be3b15e-8805-4145-87e5-40193e6921e2","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"6e49b94b-406d-48b2-b99b-d7b4d2e5e018","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"cf9de885-9548-451d-b5fc-0d133c94c5d3","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"d1844edc-4a78-4bae-86e8-b2ed9e8295b9","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"d2aa5ca0-f062-4901-a89c-2afcec57d15e","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"fa58538e-223c-45f9-acfa-0e88eb71ec83","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"36ff5ffd-5e92-4a14-acbf-aa56d5684f84","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"5948e658-ca9a-4843-8f6c-818f0212f8fa","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"5c43ec0e-a03b-449f-a2c0-301ca0532efc","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"691fcfa7-8149-4d19-977d-1df2501252ac","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"99447d6d-aed8-474f-b6d1-01a7abc0d1b0","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"b1c78b39-7ff3-4d13-a0c8-6383e71a97bd","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"d6323517-5695-4fb6-8567-391d812944e3","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"eff92c51-143d-42c5-b251-84d5e942c301","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"fc34957f-5d74-4458-8dec-afa003423b8a","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"7f1b3d00-0054-49da-9554-dda44ff2e4eb","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"660356b0-abd6-4b06-b3f5-32c9c953e8db","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"921d4878-f4dc-4f36-8e5a-813cf59d7070","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"80eabdc6-7592-4ea8-8523-9a1de827dad7","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"34c14ad2-c7d5-4a7a-a6a7-c40f83d80287","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"05cf40a3-883b-44db-9b31-0a14cf7d5075","difficultyBand":"medium","calculatorUse":null,"totalMarks":12,"questionType":"LA","isCalculation":true},{"id":"06db6322-cf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Sequences and series","topicId":"f584d5b0-d534-441f-abbf-3eee6ac58e70","topicSidebar":"$L3c","topicSlug":"sequences-and-series","topicTitle":"1.3 Sequences and series"}]}],"$L3d","$L3e"]}]

1.3 Sequences and series

Question 9

1.3 Sequences and series Questions

1.3 Sequences and series

Question 9

1.3 Sequences and series Questions