## - Extension of the Binomial Expansion

For any real number $$n$$

$$\displaystyle (1+x)^n=1+nx+{n(n-1)\over2!}x^2+{n(n-1)(n-2)\over3!}x^3+⋯$$
provided that $$-1<x<1$$
$$\displaystyle (a+x)^n=\left[a\left(1+{x\over a}\right)\right]^n$$
$$\displaystyle =a^n\left(1+{x\over a}\right)^n$$
$$\displaystyle =a^n\left[1+n\left({x\over a}\right)+{n(n-1)\over 2!} \left({x\over a}\right)^2+{n(n-1)(n-2)\over3!}\left({x\over a}\right)^3+⋯\right]$$
$$\displaystyle -1<{x\over a}<1 → -a<x<a$$

LESSON 1
Find the binomial expansion of $$\sqrt{(1+2x)^3}$$ up to and including the term in $$x^3$$.
SOLUTION
• Rewrite using indices
$$\displaystyle \sqrt{(1+2x)^3}=(1+2x)^{{3\over2}}$$
• Substitute $$\displaystyle n={3\over2}$$ and $$2x=x$$ into
$$\displaystyle (1+x)^n=1+nx+{n(n-1)\over2!}x^2+{n(n-1)(n-2)\over3!}x^3+⋯$$
$$\displaystyle (1+2x)^{{3\over2}}=1+\left({3\over2}\right)(2x)+\left({3\over2}\right)\left({3\over2}-1\right)\left({1\over2(1)}\right)(2x)^2+\left({3\over2}\right)\left({3\over2}-1\right)\left({3\over2}-2\right)\left({1\over 3(2)(1)}\right)(2x)^3$$
$$\displaystyle ⟹1+3x+\left({3\over2}\right)\left({1\over2}\right)\left({1\over2}\right)4x^2+\left({3\over2}\right)\left({1\over2}\right)\left(-{1\over2}\right)\left({1\over (3)(2)}\right)8x^3$$
$$\displaystyle ⟹1+3x+{3\over2} x^2-{1\over2} x^3$$
Valid when
$$-1<2x<1 \to -{1\over2}<x<{1\over2}$$

LESSON 2
Find the first three terms of the expansion of $$\displaystyle {4\over 2-3x}$$.
SOLUTION
• Rewrite using indices
$$\displaystyle {4\over 2-3x}=4(2-3x)^{-1}$$
• Rewrite in the form
$$\displaystyle a^n\left(1+{x\over a}\right)^n$$
$$\displaystyle 4(2-3x)^{-1}=4\left[2\left(1-{3\over2} x\right)\right]^{-1}$$
$$\displaystyle =4\left[2^{-1}\left(1-{3\over2} x\right)^{-1}\right]$$
$$\displaystyle=2\left(1-{3\over2} x\right)^{-1}$$
• Use the formula
$$\displaystyle a^n \left[1+n\left({x\over a}\right)+{n(n-1)\over2!}\left({x\over a}\right)^2+{n(n-1)(n-2)\over3!} \left({x\over a}\right)^3+⋯\right]$$
where $$n=-1$$ and $$\displaystyle {x\over a}=-{3\over2} x$$
$$\displaystyle =2\left[1+(-1)\left(-{3\over2} x\right)+(-1)(-1-1)\left({1\over (2)(1)}\right)\left(-{3\over2} x\right)^2+⋯\right]$$
$$\displaystyle =2\left[1+{3\over2} x+{9\over4}x^2+⋯\right]$$
$$\displaystyle ⟹2+3x+{9\over2} x^2$$

LESSON 3
(i) Express $$\displaystyle {9x\over (1+x)(2+5x)}$$ in the form $$\displaystyle {A\over 1+x}+{B\over 2+5x}$$ where $$A$$ and $$B$$ are integers.
(ii) Hence, or otherwise, find the expansion of $$\displaystyle {9x\over (1+x)(2+5x)}$$ as a power series in ascending order up to and including the term in $$x^3$$.
(iii) Find the range of values of $$x$$ for which the series expansion of
$$\displaystyle {9x\over (1+x)(2+5x)}$$
SOLUTION
(i) Equate terms
$$\displaystyle {9x\over (1+x)(2+5x)}={A\over 1+x}+{B\over 2+5x}$$
$$\displaystyle ⟹{9x\over (1+x)(2+5x)}={A(2+5x)+B(1+x)\over (1+x)(2+5x)}$$
$$⟹9x=A(2+5x)+B(1+x)$$
$$⟹9x=2A+5Ax+B+Bx$$
• Equate constants:
$$2A+B=0\:\:\:\:\:(1)$$
• Equate coefficients of $$x$$:
$$5A+B=9\:\:\:\:\:(2)$$
• Solving (1) and (2) simultaneously we get
$$A=3$$
$$B=-6$$
$$\displaystyle {9x\over (1+x)(2+5x)}={3\over 1+x}-{6\over 2+5x}$$

(ii) Rewrite using indices

$$\displaystyle {3\over 1+x}=3(1+x)^{-1}$$
• Substitute $$n=-1$$ into
$$\displaystyle (1+x)^n=1+nx+{n(n-1)\over2!} x^2+{n(n-1)(n-2)\over3!} x^3+⋯$$
$$\displaystyle =3\left[1+(-1)x+(-1)(-1-1)\left({1\over(2)(1)}\right) x^2+(-1)(-1-1)(-1-2)\left({1\over(3)(2)(1)}\right) x^3\right]$$
$$\displaystyle =3[1-x+x^2-x^3 ]$$
$$=3-3x+3x^2-3x^3$$
• Rewrite in the form $$\displaystyle a^n \left(1+{x\over a}\right)^n$$
$$\displaystyle {6\over 2+5x}=6(2+5x)^{-1}$$
$$\displaystyle =6\left[2\left(1+{5\over2}x\right)\right]^{-1}$$
$$\displaystyle =6\left[2^{-1}\left(1+{5\over2}x\right)^{-1}\right]$$
$$\displaystyle =3\left(1+{5\over2}x\right)^{-1}$$
• Use the formula
$$\displaystyle a^n \left[1+n\left({x\over a}\right)+{n(n-1)\over2!}\left({x\over a}\right)^2+{n(n-1)(n-2)\over3!} \left({x\over a}\right)^3+⋯\right]$$
with $$n=-1$$ and $$\displaystyle {x\over2}={5\over2}x$$
$$\displaystyle 3\left[1+(-1)\left({5\over2}x\right)+(-1)(-1-1)\left({1\over (2)(1)}\right) \left({5\over2} x\right)^2+(-1)(-1-1)(-1-2)\left({1\over (3)(2)(1)}\right)\left({5\over2}x\right)^3\right]$$
$$\displaystyle 3\left[1-{5\over2} x+{25\over4}x^2-{125\over8} x^3 \right]$$
$$\displaystyle 3-{15\over2} x+{75\over4}x^2-{375\over8} x^3$$
• Combine the two expansions
$$\displaystyle {3\over 1+x}-{6\over 2+5x}=[3-3x+3x^2-3x^3 ]-\left[3-{15\over2} x+{75\over4} x^2-{375\over8}x^3\right]$$
$$\displaystyle ={9\over2} x-{63\over4} x^2+{351\over8}x^3$$
(iii) Valid for
$$-1<x<1$$ and $$\displaystyle -1<{5\over2} x<1$$
$$\displaystyle -{2\over5}<x<{2\over5}$$
$$\displaystyle -{2\over5}<x<{2\over5}$$

LESSON 4
Use the expansion of $$\sqrt[4]{1+x}$$ to estimate $$\sqrt[4]{82}$$ to four decimal places.
SOLUTION
• Rewrite using indices and use the appropriate formula
$$\displaystyle \sqrt[4]{1+x}=(1+x)^{{1\over4}}$$
$$\displaystyle =1+{1\over4} x+\left({1\over4}\right)\left({1\over4}-1\right)\left({1\over(2)(1)}\right) x^2+\left({1\over4}\right)\left({1\over4}-1\right)\left({1\over4}-2\right)\left({1\over (3)(2)(1)}\right)x^3$$
$$\displaystyle =1+{1\over4} x-{3\over32} x^2+{7\over128} x^3$$
• Let $$1+x=82$$ and solve for $$x$$
$$x=81$$
The formula only works if $$x$$ is small. Since 81 is large we need to rewrite as follows:
$$\displaystyle (1+81)^{{1\over4}}=\left[81\left(1+{1\over81}\right)\right]^{{1\over4}}$$
$$\displaystyle =81^{{1\over4}} \left(1+{1\over81}\right)^{{1\over4}}$$
$$\displaystyle =3\left(1+{1\over81}\right)^{{1\over4}}$$
• We now have that $$\displaystyle x={1\over81}$$ which is sufficiently small.
$$\displaystyle =3\left[1+{1\over4}\left({1\over81}\right)-{3\over32}\left({1\over81}\right)^2+{7\over128}\left({1\over81}\right)^3\right]$$
$$=3[1.003]$$
$$=3.009$$

LESSON 5
(i) Find the binomial expansion of $$\displaystyle (1+6x)^{{2\over3}}$$ up to and including the term in $$x^2$$.
(ii) Find the binomial expansion of $$\displaystyle (8+6x)^{{2\over3}}$$ up to and including the term in $$x^2$$.
(iii) Hence, find an estimate for the value of $$\sqrt[3]{144}$$ in the form $$\displaystyle {a\over b}$$ where $$a$$ and $$b$$ are integers.
SOLUTION
(a) Use the appropriate formula
$$\displaystyle (1+6x)^{{2\over3}}=1+{2\over3}(6x)+\left({2\over3}\right)\left({2\over3}-1\right) \left({1\over (2)(1)}\right)(6x)^2$$
$$=1+4x-4x^2$$
• Rewrite and use the appropriate formula
$$\displaystyle (8+6x)^{{2\over3}}=\left[8\left(1+{6\over8}x\right)\right]^{{2\over3}}$$
$$\displaystyle =8^{{2\over3}}\left(1+{6\over8}x\right)^{{2\over3}}$$
$$\displaystyle =4\left(1+6\left({1\over8}x\right)\right)^{{2\over3}}$$
• In part (a) we expanded $$\displaystyle (1+6x)^{{1\over3}}$$ to get $$1+4x-4x^2$$. We now have to expand $$\displaystyle \left(1+6\left({1\over8} x\right)\right)^{{2\over3}}$$. The appropriate formula will work but alternately we can substitute $$\displaystyle {1\over8} x$$ for $$x$$.
$$\displaystyle =4\left[1+4\left({1\over8} x\right)-4\left({1\over8}\right)^2\right]$$
$$\displaystyle =4+2x-{1\over4} x^2$$
(c) Rewrite as an expression with an index of $$\displaystyle {2\over3}$$.
$$\displaystyle (144)^{{1\over3}}=(12^2)^{{1\over3}}=12^{{2\over3}}$$
• Let $$8+6x=12$$ and solve for $$x$$
$$8+6x=12$$
$$6x=4$$
$$\displaystyle x={2\over3}$$
• Substitute $$\displaystyle x={2\over3}$$ into $$\displaystyle 4+2x-{1\over 4x^2}$$
$$\displaystyle \sqrt[3]{144}\approx 4+2\left({2\over3}\right)-{1\over4} \left({2\over3}\right)^2$$
$$\displaystyle\approx {47\over9}$$