How can you use binomial series to expand ${\left( {1 + x} \right)^n}$?
Answer
613.2k+ views
Hint: In this problem, we have to expand the expression ${\left( {1 + x} \right)^n}$ by using binomial series. As the expression has two different terms 1 and $x$. So, it is a binomial series. Using the binomial theorem, we will use the formula, by which we can expand it till ${n^{th}}$ 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 ${\left( {x + y} \right)^n}$ into a sum containing terms of the form $a{x^b}{y^c}$. According to this theorem, where the exponents b and c are non-negative integers with \[b + c = n\], 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 ${\left( {a + b} \right)^n}$ can be expressed as,
${\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,
\[{\left( {a + b} \right)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + \ldots + {}^n{C_n}{a^0}{b^n}\]
As we know,
$ \Rightarrow {}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$
So, the binomial expansion can be written as,
$ \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, $a = 1,b = x$. Substitute the values in the above expansion,
$ \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,
$ \Rightarrow {\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{2}{x^2} + \ldots + {x^n}$
Hence, the expansion of ${\left( {1 + x} \right)^n}$ is $1 + nx + \dfrac{{n\left( {n - 1} \right)}}{2}{x^2} + \ldots + {x^n}$.
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 ${\left( {a + b} \right)^n}$ the binomial expansion is \[\left( {n + 1} \right)\]. 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.
Complete step by step answer:
The binomial theorem determines the algebraic expansion of a binomial's powers. We can extend the polynomial ${\left( {x + y} \right)^n}$ into a sum containing terms of the form $a{x^b}{y^c}$. According to this theorem, where the exponents b and c are non-negative integers with \[b + c = n\], 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 ${\left( {a + b} \right)^n}$ can be expressed as,
${\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,
\[{\left( {a + b} \right)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + \ldots + {}^n{C_n}{a^0}{b^n}\]
As we know,
$ \Rightarrow {}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$
So, the binomial expansion can be written as,
$ \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, $a = 1,b = x$. Substitute the values in the above expansion,
$ \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,
$ \Rightarrow {\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{2}{x^2} + \ldots + {x^n}$
Hence, the expansion of ${\left( {1 + x} \right)^n}$ is $1 + nx + \dfrac{{n\left( {n - 1} \right)}}{2}{x^2} + \ldots + {x^n}$.
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 ${\left( {a + b} \right)^n}$ the binomial expansion is \[\left( {n + 1} \right)\]. 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.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

