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Question: How do you use the binomial series to expand \[\dfrac{1}{{{\left( 2+x \right)}^{3}}}\]?...

How do you use the binomial series to expand 1(2+x)3\dfrac{1}{{{\left( 2+x \right)}^{3}}}?

Explanation

Solution

In this problem, we have to use the binomial series to expand the given expression. We know that the binomial series is a kind of formula that helps us to expand binomials raised to the power of any number using the binomial theorem. The binomial series formula is (a+b)n=an+nan1b+n(n1)1×2an2b2+n(n1)(n2)1×2×3an3b3...{{\left( a+b \right)}^{n}}={{a}^{n}}+n{{a}^{n-1}}b+\dfrac{n\left( n-1 \right)}{1\times 2}{{a}^{n-2}}{{b}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{1\times 2\times 3}{{a}^{n-3}}{{b}^{3}}.... We can write the given expression in terms of the right-hand side of the binomial theorem and we can substitute the values, a, b. n in the binomial series formula to expand it.

Complete step by step answer:
We have to use the binomial theorem to expand 1(2+x)3\dfrac{1}{{{\left( 2+x \right)}^{3}}}.
We know that the binomial series is a kind of formula that helps us to expand binomials raised to the power of any number using the binomial theorem.
We know that the binomial series is,
(a+b)n=an+nan1b+n(n1)1×2an2b2+n(n1)(n2)1×2×3an3b3...{{\left( a+b \right)}^{n}}={{a}^{n}}+n{{a}^{n-1}}b+\dfrac{n\left( n-1 \right)}{1\times 2}{{a}^{n-2}}{{b}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{1\times 2\times 3}{{a}^{n-3}}{{b}^{3}}...
We can write the given expression as,
1(2+x)3=(2+x)3\Rightarrow \dfrac{1}{{{\left( 2+x \right)}^{-3}}}={{\left( 2+x \right)}^{-3}}
Here, a = 2, b = x, n = -3.
We can now expand the expression using binomial series, we get
(2+x)3=23+(3)(24)x+3(4)1×225x2+3(4)(5)626x3...\Rightarrow {{\left( 2+x \right)}^{-3}}={{2}^{-3}}+\left( -3 \right)\left( {{2}^{-4}} \right)x+\dfrac{-3\left( -4 \right)}{1\times 2}{{2}^{-5}}{{x}^{2}}+\dfrac{-3\left( -4 \right)\left( -5 \right)}{6}{{2}^{-6}}{{x}^{3}}...
We can now simplify the above step, we get
1(2+x)3=18316x+316x2532x3+...\Rightarrow \dfrac{1}{{{\left( 2+x \right)}^{3}}}=\dfrac{1}{8}-\dfrac{3}{16}x+\dfrac{3}{16}{{x}^{2}}-\dfrac{5}{32}{{x}^{3}}+...

Therefore, the expansion using binomial series is, 1(2+x)3=18316x+316x2532x3+...\dfrac{1}{{{\left( 2+x \right)}^{3}}}=\dfrac{1}{8}-\dfrac{3}{16}x+\dfrac{3}{16}{{x}^{2}}-\dfrac{5}{32}{{x}^{3}}+....

Note: we should know that, any number with the negative power, can be written in a fraction form, by writing it in the denominator with the same power as positive. We should remember that the binomial series is a kind of formula that helps us to expand binomials raised to the power of any number using the binomial theorem.