ty 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."},{"t

1.3 Sequences and series

Question 6

1.3 Sequences and series Questions

1.3 Sequences and series

Question 6

1.3 Sequences and series Questions