To expand $(1 - 4x)^{\frac{1}{4}}$ up to and including the term in $x^2$, we use the binomial expansion formula for fractional powers:

$(1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \dots$ Here:

• $n = \frac{1}{4}$ (the power),
• $u = -4x$ (the term inside the bracket after factoring out 1).

Step 1: First few terms of the expansion

Substitute $n = \frac{1}{4}$ and $u = -4x$:

1. Term 1: The first term is simply $1$.

2. Term 2: The second term is $n \cdot u$: $\frac{1}{4} \cdot (-4x) = -x.$

3. Term 3: The third term is $\frac{n(n-1)}{2!} \cdot u^2$:

   • First, calculate $n(n-1)$: $\frac{1}{4} \left( \frac{1}{4} - 1 \right) = \frac{1}{4} \left( -\frac{3}{4} \right) = -\frac{3}{16}.$
   • Now, divide by $2! = 2$: $\frac{-\frac{3}{16}}{2} = -\frac{3}{32}.$
   • Multiply by $u^2 = (-4x)^2 = 16x^2$: $-\frac{3}{32} \cdot 16x^2 = -\frac{48}{32}x^2 = -\frac{3}{2}x^2.$

Step 2: Combine terms

Now, collect the terms up to $x^2$: $(1 - 4x)^{\frac{1}{4}} = 1 - x - \frac{3}{2}x^2 + \dots$

Final Answer:

The binomial expansion of $(1 - 4x)^{\frac{1}{4}}$ up to and including the term in $x^2$ is: $1 - x - \frac{3}{2}x^2.$

Do you need further clarification or additional steps? Let me know!

Here are 5 related questions for further practice:

1. Expand $(1 + 3x)^{\frac{1}{3}}$ up to the $x^2$ term.
2. Use the binomial theorem to find the expansion of $(1 - 2x)^4$.
3. Determine the coefficient of $x^3$ in the expansion of $(1 + 2x)^{\frac{1}{2}}$.
4. Write the first four terms of $(1 - 5x)^{\frac{2}{3}}$ using the binomial series.
5. Prove that the binomial expansion of $(1 + u)^n$ converges for $|u| < 1$.

Tip: For fractional powers, the binomial expansion is infinite, but you only need to include terms up to the required power!