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Question

Question: How do you use the binomial series to expand \[{{\left( 1+2x \right)}^{5}}\]?...

How do you use the binomial series to expand (1+2x)5{{\left( 1+2x \right)}^{5}}?

Explanation

Solution

The term of the form (a+b)n{{\left( a+b \right)}^{n}} is called a binomial term. The general form of the expansion of the binomial term is r=0nnCrarbnr\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}}. We can find the binomial series by substituting the values of a, b, and n in this summation. It should be noted that here nn is a positive integer.

Complete step-by-step answer:
We are asked to expand the binomial term (1+2x)5{{\left( 1+2x \right)}^{5}}. Comparing with the general binomial term (a+b)n{{\left( a+b \right)}^{n}}, we get a=1,b=2x&n=5a=1,b=2x\And n=5. We know the general form of the binomial series is r=0nnCrarbnr\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}}. We can expand the given binomial term by substituting the value of variables in the general form, as follows
r=0nnCrarbnr\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}}
r=055Cr(1)r(2x)5r\Rightarrow \sum\limits_{r=0}^{5}{^{5}{{C}_{r}}{{\left( 1 \right)}^{r}}{{\left( 2x \right)}^{5-r}}}
Expanding the above summation, we get
5C0(1)0(2x)50+5C1(1)1(2x)51+5C2(1)2(2x)52+5C3(1)3(2x)53+5C4(1)4(2x)54+5C5(1)5(2x)55{{\Rightarrow }^{5}}{{C}_{0}}{{\left( 1 \right)}^{0}}{{\left( 2x \right)}^{5-0}}{{+}^{5}}{{C}_{1}}{{\left( 1 \right)}^{1}}{{\left( 2x \right)}^{5-1}}{{+}^{5}}{{C}_{2}}{{\left( 1 \right)}^{2}}{{\left( 2x \right)}^{5-2}}{{+}^{5}}{{C}_{3}}{{\left( 1 \right)}^{3}}{{\left( 2x \right)}^{5-3}}{{+}^{5}}{{C}_{4}}{{\left( 1 \right)}^{4}}{{\left( 2x \right)}^{5-4}}{{+}^{5}}{{C}_{5}}{{\left( 1 \right)}^{5}}{{\left( 2x \right)}^{5-5}}
We know that nCr=n!r!(nr)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}, using this to simplify the above series we get
5C0=5!0!(50)!=1^{5}{{C}_{0}}=\dfrac{5!}{0!\left( 5-0 \right)!}=1
5C1=5!1!(51)!=5^{5}{{C}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!}=5
5C2=5!2!(52)!=10^{5}{{C}_{2}}=\dfrac{5!}{2!\left( 5-2 \right)!}=10
5C3=5!3!(53)!=10^{5}{{C}_{3}}=\dfrac{5!}{3!\left( 5-3 \right)!}=10
5C4=5!4!(54)!=5^{5}{{C}_{4}}=\dfrac{5!}{4!\left( 5-4 \right)!}=5
5C5=5!5!(55)!=1^{5}{{C}_{5}}=\dfrac{5!}{5!\left( 5-5 \right)!}=1
Substituting the values of the above coefficients, we get
1(2x)5+5(2x)4+10(2x)3+10(2x)2+5(2x)1+1(2x)0\Rightarrow 1{{\left( 2x \right)}^{5}}+5{{\left( 2x \right)}^{4}}+10{{\left( 2x \right)}^{3}}+10{{\left( 2x \right)}^{2}}+5{{\left( 2x \right)}^{1}}+1{{\left( 2x \right)}^{0}}
Simplifying the above series, we get
32x5+80x4+80x3+40x2+10x+1\Rightarrow 32{{x}^{5}}+80{{x}^{4}}+80{{x}^{3}}+40{{x}^{2}}+10x+1
Is the expansion of the given binomial term.

Note: We can use more special binomial expansions to expand the series. Here one of the terms inside the bracket is 1. Hence, we can use the expansion of (1+x)n{{\left( 1+x \right)}^{n}} whose general form of expansion is r=0nnCrxr\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{r}}}. For this series, we have to substitute 2x2x at the place of xx. These expansions are very important and should be remembered.
We can use these expansions only when nn is a positive integer. For cases when the nn is a non-positive integer, we need to use different types of expansions.