ifies to\n$$\\left(1+\\frac{2}{3}x\\right)^{-\\frac{1}{2}}=1-\\frac{x}{3}+\\frac{x^2}{6}-\\frac{5x^3}{54}+\\cdots$$\nMultiplying by $\\frac{1}{3}$ gives\n$$(9+6x)^{-\\frac{1}{2}}=\\frac{1}{3}-\\frac{x}{9}+\\frac{x^2}{18}-\\frac{5x^3}{162}+\\cdots$$"},{"id":"block-3","content":"The expansion is valid while $|u|<1$. Here that means\n$$\\left|\\frac{2}{3}x\\right|<1 \\Rightarrow |x|<\\frac{3}{2}$$\nso the range of validity is $-\\frac{3}{2}-\\frac{3}{2}$","isCorrect":false},{"id":"d","text":"All real values of $x$","isCorrect":false}],"question":"What is the range of validity for the expansion of $(9+6x)^{-\\frac{1}{2}}$?","explanation":"After factoring, the bracket is $\\left(1+\\frac{2}{3}x\\right)^{-\\frac{1}{2}}$, so $u=\\frac{2}{3}x$. For fractional or negative powers we need $|u|<1$, giving $\\left|\\frac{2}{3}x\\right|<1$ and therefore $|x|<\\frac{3}{2}$. The common wrong answer B comes from not solving the inequality fully for $x$.","correctOptionId":"a"},{"id":"card-9","type":"flashcard","cards":[{"id":"fc-1","back":"One part of a sum, such as $6x^2$.","front":"Term"},{"id":"fc-2","back":"The number multiplying a power of the variable, such as 6 in $6x^2$.","front":"Coefficient"},{"id":"fc-3","back":"Constant term first, then $x$, then $x^2$, then $x^3$, and so on.","front":"Ascending powers of $x$"},{"id":"fc-4","back":"Rewrite the expression as $A(1+u)^n$ before expanding.","front":"Standard form for the infinite formula"},{"id":"fc-5","back":"For fractional or negative powers, the expansion is valid when $|u|<1$.","front":"Validity condition"}],"order":9},{"id":"card-10","type":"content","order":10,"title":"Worked example: approximation","blocks":[{"id":"block-1","content":"Binomial expansions are useful when $|u|$ is small, because higher powers get tiny very quickly. To estimate $\\sqrt{8.8}$, we set $9-2x=8.8$, so $x=0.1$ and $\\sqrt{8.8}=(9-2x)^{\\frac{1}{2}}$."},{"id":"block-2","content":"Now factor the 9 and expand the expression using the binomial formula:\n$$\n(9-2x)^{\\frac{1}{2}}=3\\left(1-\\frac{2x}{9}\\right)^{\\frac{1}{2}}\n$$\nUsing the general expansion $(1+u)^{\\frac{1}{2}}=1+\\frac{1}{2}u-\\frac{1}{8}u^2+\\frac{1}{16}u^3+\\cdots$, we get\n$$\n(9-2x)^{\\frac{1}{2}}=3-\\frac{x}{3}-\\frac{x^2}{54}-\\frac{x^3}{486}+\\cdots\n$$"},{"id":"block-3","content":"Substitute $x=0.1$ into the series to find the numerical estimate:\n$$\n3-\\frac{0.1}{3}-\\frac{0.1^2}{54}-\\frac{0.1^3}{486}=2.966479\\ldots\n$$\nSo $\\sqrt{8.8}\\approx 2.96648$ to 5 decimal places. Because $|u|=\\frac{1}{45}$ is very small, the approximation is highly accurate."}]},{"id":"card-11","type":"fill_blank","order":11,"blanks":[{"id":"approx","isMath":true,"correctAnswer":"2.96648"}],"sentence":"Using $x=0.1$ in the expansion of $(9-2x)^{\\frac{1}{2}}$ gives $\\sqrt{8.8} \\approx$ {{blank:approx}} to 5 decimal places."},{"id":"card-12","type":"content","order":12,"title":"Worked example: multiplying expansions","blocks":[{"id":"block-1","content":"Sometimes you expand one part and then multiply by another factor. For\n$$\\frac{1+x}{(1-2x)^2}=(1+x)(1-2x)^{-2}$$\nexpand the term $(1-2x)^{-2}$ first using the binomial series formula."},{"id":"block-2","content":"Using $n=-2$ and $u=-2x$, the expansion is:\n$$(1-2x)^{-2}=1+4x+12x^2+32x^3+\\cdots$$"},{"id":"block-3","content":"Now multiply this series by $1+x$, ensuring you only keep terms that contribute up to $x^3$.\n$$(1+x)(1+4x+12x^2+32x^3)=1+4x+12x^2+32x^3+x+4x^2+12x^3+\\cdots$$"},{"id":"block-4","content":"By collecting like terms, we get the final result:\n$$\\frac{1+x}{(1-2x)^2}=1+5x+16x^2+44x^3+\\cdots$$\nAny products resulting in $x^4$ or higher powers can be ignored for this specific expansion."}]},{"id":"card-13","type":"mcq","order":13,"options":[{"id":"a","text":"$$a=4,\\ b=\\frac{5}{2}$","isCorrect":true},{"id":"b","text":"$$a=2,\\ b=1$","isCorrect

Binomial Expansion Lesson

Binomial Expansion Lesson