
How do you use the binomial series to expand \[\dfrac{1}{{{\left( 2+x \right)}^{3}}}\]?
Answer
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Hint: 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 \[{{\left( a+b \right)}^{n}}={{a}^{n}}+n{{a}^{n-1}}b+\dfrac{n\left( n-1 \right)}{1\times 2}{{a}^{n-2}}{{b}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{1\times 2\times 3}{{a}^{n-3}}{{b}^{3}}...\]. 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 \[\dfrac{1}{{{\left( 2+x \right)}^{3}}}\].
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,
\[{{\left( a+b \right)}^{n}}={{a}^{n}}+n{{a}^{n-1}}b+\dfrac{n\left( n-1 \right)}{1\times 2}{{a}^{n-2}}{{b}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{1\times 2\times 3}{{a}^{n-3}}{{b}^{3}}...\]
We can write the given expression as,
\[\Rightarrow \dfrac{1}{{{\left( 2+x \right)}^{-3}}}={{\left( 2+x \right)}^{-3}}\]
Here, a = 2, b = x, n = -3.
We can now expand the expression using binomial series, we get
\[\Rightarrow {{\left( 2+x \right)}^{-3}}={{2}^{-3}}+\left( -3 \right)\left( {{2}^{-4}} \right)x+\dfrac{-3\left( -4 \right)}{1\times 2}{{2}^{-5}}{{x}^{2}}+\dfrac{-3\left( -4 \right)\left( -5 \right)}{6}{{2}^{-6}}{{x}^{3}}...\]
We can now simplify the above step, we get
\[\Rightarrow \dfrac{1}{{{\left( 2+x \right)}^{3}}}=\dfrac{1}{8}-\dfrac{3}{16}x+\dfrac{3}{16}{{x}^{2}}-\dfrac{5}{32}{{x}^{3}}+...\]
Therefore, the expansion using binomial series is, \[\dfrac{1}{{{\left( 2+x \right)}^{3}}}=\dfrac{1}{8}-\dfrac{3}{16}x+\dfrac{3}{16}{{x}^{2}}-\dfrac{5}{32}{{x}^{3}}+...\].
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.
Complete step by step answer:
We have to use the binomial theorem to expand \[\dfrac{1}{{{\left( 2+x \right)}^{3}}}\].
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,
\[{{\left( a+b \right)}^{n}}={{a}^{n}}+n{{a}^{n-1}}b+\dfrac{n\left( n-1 \right)}{1\times 2}{{a}^{n-2}}{{b}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{1\times 2\times 3}{{a}^{n-3}}{{b}^{3}}...\]
We can write the given expression as,
\[\Rightarrow \dfrac{1}{{{\left( 2+x \right)}^{-3}}}={{\left( 2+x \right)}^{-3}}\]
Here, a = 2, b = x, n = -3.
We can now expand the expression using binomial series, we get
\[\Rightarrow {{\left( 2+x \right)}^{-3}}={{2}^{-3}}+\left( -3 \right)\left( {{2}^{-4}} \right)x+\dfrac{-3\left( -4 \right)}{1\times 2}{{2}^{-5}}{{x}^{2}}+\dfrac{-3\left( -4 \right)\left( -5 \right)}{6}{{2}^{-6}}{{x}^{3}}...\]
We can now simplify the above step, we get
\[\Rightarrow \dfrac{1}{{{\left( 2+x \right)}^{3}}}=\dfrac{1}{8}-\dfrac{3}{16}x+\dfrac{3}{16}{{x}^{2}}-\dfrac{5}{32}{{x}^{3}}+...\]
Therefore, the expansion using binomial series is, \[\dfrac{1}{{{\left( 2+x \right)}^{3}}}=\dfrac{1}{8}-\dfrac{3}{16}x+\dfrac{3}{16}{{x}^{2}}-\dfrac{5}{32}{{x}^{3}}+...\].
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.
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