- 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 \(\displaystyle 4+2x-{1\over 4x^2}\)
\(\displaystyle \sqrt[3]{144}\approx 4+2\left({2\over3}\right)-{1\over4} \left({2\over3}\right)^2\)
\(\displaystyle\approx {47\over9}\)