Use the binomial theorem to expand $ ext{ } ext { } ext{ } ext{ } (4-9x)^{rac{1}{2}} $ $|x| < rac{4}{9}$, in ascending powers of $x$, up to and including the term in $x^{3}$, simplifying each term. - Edexcel - A-Level Maths Pure - Question 3 - 2005 - Paper 6

We begin by identifying the binomial expansion formula:

(a + b)^n = inom{n}{0} a^n b^0 + inom{n}{1} a^{n-1} b + inom{n}{2} a^{n-2} b^2 + inom{n}{3} a^{n-3} b^3 + ...

For our case, set:

• $a = 4$,
• $b = -9x$,
• n = rac{1}{2}.

Using the binomial theorem, we get:

(4 - 9x)^{rac{1}{2}} = 4^{rac{1}{2}} + inom{rac{1}{2}}{1} 4^{rac{1}{2}-1} (-9x) + inom{rac{1}{2}}{2} 4^{rac{1}{2}-2} (-9x)^2 + inom{rac{1}{2}}{3} 4^{rac{1}{2}-3} (-9x)^3 + ...

Calculating the terms:

• First term: 4^{rac{1}{2}} = 2.
• Second term: inom{rac{1}{2}}{1} = rac{1}{2}; ext{ so, } rac{1}{2} imes 2^{-1} imes (-9x) = -rac{9x}{4}.
• Third term: inom{rac{1}{2}}{2} = rac{rac{1}{2} imes rac{-rac{1}{2}}{2}}{2!} = -rac{1}{8}; ext{ so, } -rac{1}{8} imes 4^{-1} imes 81x^2 = -rac{81}{32} x^2.
• Fourth term: inom{rac{1}{2}}{3} = rac{rac{1}{2} imes (-rac{1}{2}) imes (-rac{3}{2})}{3!} = rac{1}{16}; ext{ so, } rac{1}{16} imes 4^{-rac{3}{2}} imes (-729x^3) = -rac{729}{512} x^3.

Combining these, we have:

(4-9x)^{rac{1}{2}} = 2 - rac{9x}{4} - rac{81}{32} x^2 - rac{729}{512} x^3 + ...