Cards In This Deck

• $$\begin{gathered} \text{Does the expansion of }(1+5x{)}^n\text{ produce a finite or infinite number of terms?} \\ \text{Scenario A: }n=3 \\ \text{Scenario B: }n=-3 \end{gathered}$$
  $$\begin{gathered} \text{Scenario A: Finite (stops after }{x}^3\text{ term).} \\ \text{Scenario B: Infinite.} \end{gathered}$$
• $\text{What is the general term formula for the coefficient of }{x}^2\text{ in the expansion of }(1+x{)}^n\text{?}$
  $\frac{n(n-1)}{2!}$
• $$\begin{gathered} \text{True or False?} \\ \text{The expansion of }(1-3x{)}^{-2}\text{ involves terms like }\frac{(-2)(-3)}{2!}(3x{)}^2\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Term:} \\ \frac{(-2)(-3)}{2!}(-3x{)}^2 \\ \text{The }x\text{ in the formula must be replaced by the entire term including the sign. }(-3x{)}^2\text{ becomes positive }9x^2\text{, but failing to write the negative sign in the bracket often leads to sign errors in odd powers (like }{x}^3\text{).} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{When finding an approximate value for }\sqrt{10}\text{ using the expansion of }(1+ax{)}^n\text{, you should equate }1+ax=10\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False (Usually)} \\ \text{Setting }1+ax=10\text{ implies }ax=9\text{. If }x\text{ is large, the approximation will be inaccurate or the series may not converge (if }|ax|\ge 1\text{).} \\ \text{✅ Better Strategy:} \\ \text{Factor out a square number close to 10 (e.g., 9).} \\ \sqrt{10}=\sqrt{9\times \frac{10}{9}}=3(1+\frac{1}{9}{)}^{\frac{1}{2}} \\ \text{Then solve }1+ax=1+\frac{1}{9}\text{, so }x\text{ is small.} \end{gathered}$$
• $\text{State the range of values for which the binomial expansion of }(1+bx{)}^n\text{ is valid, where }n\text{ is a fraction or negative integer.}$
  $$\begin{gathered} |bx|<1 \\ \text{OR} \\ |x|<\frac{1}{|b|} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{The expansion of }(a+bx{)}^n\text{ is valid for }|\frac{b}{a}x|\le 1\text{ when }n\text{ is negative.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{It is strictly less than 1.} \\ |\frac{b}{a}x|<1 \\ \text{It does not include the "equals" case (unlike geometric series convergence which is also strictly <1).} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{To expand }(4+2x{)}^{\frac{1}{2}}\text{, you can immediately substitute }n=\frac{1}{2}\text{ and }x=2x\text{ into the standard binomial formula.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{The standard formula }(1+x{)}^n\text{ requires the first term to be 1.} \\ \text{✅ Correct Step:} \\ \text{Factor out the 4 first:} \\ (4+2x{)}^{\frac{1}{2}}=4^{\frac{1}{2}}(1+\frac{2x}{4}{)}^{\frac{1}{2}}=2(1+\frac{x}{2}{)}^{\frac{1}{2}} \\ \text{Then expand }(1+\frac{x}{2}{)}^{\frac{1}{2}}\text{.} \end{gathered}$$
• $\text{How do you rewrite the expression }\frac{3}{2-x}\text{ to prepare it for binomial expansion?}$
  $$\begin{gathered} \text{1. Write in index form: }3(2-x{)}^{-1} \\ \text{2. Factor out the constant from the bracket to get a ’1’:} \\ 3[2(1-\frac{x}{2}){]}^{-1} \\ 3(2{)}^{-1}(1-\frac{x}{2}{)}^{-1} \\ =\frac{3}{2}(1-\frac{x}{2}{)}^{-1} \end{gathered}$$
• $$\begin{gathered} \text{You have expanded }f(x)\text{ using partial fractions. } \\ \text{Part A is valid for }|x|<1\text{. } \\ \text{Part B is valid for }|x|<2\text{. } \\ \text{What is the validity range for the whole expansion?} \end{gathered}$$
  $|x|<1$
• $\text{What is the first step to expand }g(x)=\frac{4-5x}{(1+x)(2-x)}\text{ using the binomial theorem?}$
  $$\begin{gathered} \text{1. Express }g(x)\text{ as partial fractions.} \\ \text{2. Rewrite the partial fractions in index form, e.g., }A(1+x{)}^{-1}+B(2-x{)}^{-1}\text{.} \\ \text{3. Expand each term separately.} \end{gathered}$$