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Question: How can you use binomial series to expand \({\left( {1 + x} \right)^n}\)?...

How can you use binomial series to expand (1+x)n{\left( {1 + x} \right)^n}?

Explanation

Solution

In this problem, we have to expand the expression (1+x)n{\left( {1 + x} \right)^n} by using binomial series. As the expression has two different terms 1 and xx. So, it is a binomial series. Using the binomial theorem, we will use the formula, by which we can expand it till nth{n^{th}} term. Then substitute the value and evaluate it to get the desired result.

Complete step by step answer:
The binomial theorem determines the algebraic expansion of a binomial's powers. We can extend the polynomial (x+y)n{\left( {x + y} \right)^n} into a sum containing terms of the form axbyca{x^b}{y^c}. According to this theorem, where the exponents b and c are non-negative integers with b+c=nb + c = n, and each term's coefficient an is a particular positive integer depending on n and b.
First of all, we need to express the general binomial expression theorem. Then we need to superimpose into that form the expression given. To obtain the desired polynomial form, we need to define the various variables and position the values.
The binomial term of the expression (a+b)n{\left( {a + b} \right)^n} can be expressed as,
(a+b)n=∑r=0nnCk⋅(xn−kyk){\left( {a + b} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_k} \cdot \left( {{x^{n - k}}{y^k}} \right)}
It can also be written as,
(a+b)n=nC0anb0+nC1an−1b1+nC2an−2b2+…+nCna0bn{\left( {a + b} \right)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + \ldots + {}^n{C_n}{a^0}{b^n}
As we know,
⇒nCr=n!(n−r)!r!\Rightarrow {}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}
So, the binomial expansion can be written as,
⇒(a+b)n=an+nan−1b+n(n−1)2an−2b2+…+bn\Rightarrow {\left( {a + b} \right)^n} = {a^n} + n{a^{n - 1}}b + \dfrac{{n\left( {n - 1} \right)}}{2}{a^{n - 2}}{b^2} + \ldots + {b^n}
According to the question, a=1,b=xa = 1,b = x. Substitute the values in the above expansion,
⇒(1+x)n=1n+n×1n−1x+n(n−1)2×1n−2x2+…+xn\Rightarrow {\left( {1 + x} \right)^n} = {1^n} + n \times {1^{n - 1}}x + \dfrac{{n\left( {n - 1} \right)}}{2} \times {1^{n - 2}}{x^2} + \ldots + {x^n}
Simplify the terms,
⇒(1+x)n=1+nx+n(n−1)2x2+…+xn\Rightarrow {\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{2}{x^2} + \ldots + {x^n}

Hence, the expansion of (1+x)n{\left( {1 + x} \right)^n} is 1+nx+n(n−1)2x2+…+xn1 + nx + \dfrac{{n\left( {n - 1} \right)}}{2}{x^2} + \ldots + {x^n}.

Note: The binomial formula is true for all positive integer values of n based on the binomial theorem's properties. The sum number of terms in (a+b)n{\left( {a + b} \right)^n} the binomial expansion is (n+1)\left( {n + 1} \right). It should also be noticed that n is the sum of the exponents of a and b in each expansion term. We will note that the exponents of a increase by 1 when expanding the binomial coefficient, while the exponents of b decrease by 1.