Expanding \((1-x)^n\)
The Binomial Infinite Series
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\[1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3\dots\]
- This works for positive, negative or fractional powers, unlike the \(nCr\) method.
Expand: \(\frac{1}{x-1}\)
- \(\frac{1}{1-x}=(1-x)^{-1}\)
- \(1+(1)(-x)+\frac{-1)(-2}{2!}(-x)^2\dots\)
- \(1-x+x^2-x^3+\dots\)
Find the binomial expansion for these terms up to \(x^3\)
- \((1+x)^{-4}=1+(-4)(x)+\frac{(-4)(-5)}{2!}x^2+\frac{(-4)(-5)(-6)}{3!}x^3\)
\(1-4x+10x^2-20x^3\)
\(|x|<1\) - \((1+x)^{-6}=1+(-6)(x)+\frac{(-6)(-7)}{2!}x^2+\frac{(-6)(-7)(-8)}{3!}x^3\)
\(1-6x+21x^2-56x^3\)
\(|x|<1\) - \((1+x)^{\frac{1}{2}}=1+\left( \frac{1}{2} \right)(x)+\frac{\left( \frac{1}{2} \right)\left( -\frac{1}{2} \right)}{2!}x^2+\frac{\left( \frac{1}{2} \right)\left( -\frac{1}{2} \right)\left( -\frac{3}{2} \right)}{3!}x^3\)
\(1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3\)
\(|x|<1\)
Find the binomial expansion for these terms up to \(x^3\)
- \((1+3x)^{-3}=1+(-3)(3x)+\frac{(-3)(-4)}{2}(3x)^2+\frac{(-3)(-4)(-5)}{6}(3x)^3\)
\(=1-9x+54x^2-270x^3\)
\(|x|<\frac{1}{3}\)
Checking for Validity
- Expansions of \((1+bx)^n\) are valid for \(|bx|<1\) or \(|x|< \frac{1}{|b|}\)
\(\frac{1+x}{1-2x}\)
a)
- \((1+x)(1-2x)^{-1}\)
- \((1-2x)^{-1}=1+(-1)(-2x)+\frac{(-1)(-2)}{2}(-2x)^2+\frac{(-1)(-2)(-3)}{6}(-2x)^{3}\)
- \(1+2x+4x^2+8x^3\)
- \((1+x)(1+2x+4x^2+8x^3)\)
- \((1+2x+4x^2+8x^3)+(x+2x^2+4x^3+8x^4)\)
- \((1+3x+6x^2+12x^3+8x^4)\)
b)
- Validity is: \(|-2x|<1\)
- Validity is: \(|x|<-\frac{1}{2}\)
In the expansion of \((1+ax)^{-\frac{1}{2}}\), the coefficient of \(x^2\) is 24. Find the possible values of \(a\), then find the coefficients of \(x^3\).
a)
- \(1+(ax)\left( -\frac{1}{2} \right)+\frac{\left( -\frac{1}{2} \right)\left( -\frac{3}{2} \right)}{2}(ax)^2+\frac{\left( -\frac{1}{2} \right)\left( -\frac{3}{2} \right)\left( -\frac{5}{2} \right)}{6}(ax)^3\)
- \(1-\frac{ax}{2}+\frac{3}{8}a^2x^2-\frac{5}{16}a^3x^3\)
- \(\frac{3}{8}a^2x^2=24x^2\)
- \(\frac{3}{8}a^2=24\)
- \(a^2=64\)
- \(a=8\)
b)
- \(a=8\)
- \(-\frac{5}{16}a^3x^3=-160x^3\)