
What is the expansion of \[{{(x+1)}^{4}}\] ?
Answer
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Hint:A Binomial expression is the equation with two terms. The binomial theorem is the algebraic expansion of powers of a binomial. According to the theorem, expansion of \[{{(x+y)}^{n}}\]is possible. The Binomial Coefficient is given by \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}\]
The Binomial Theorem is given by \[{{(x+y)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}y+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+...+{}^{n}{{C}_{n}}{{y}^{n}}\]
The equations that contain binomials are called binomial equations. Operations like addition, subtraction, multiplication, division, etc can be performed on Binomial expression.
Complete step by step answer:
The Binomial expansion of \[{{(x+1)}^{4}}\] according to the Binomial Theorem can be written as
\[{{(x+y)}^{4}}={}^{4}{{C}_{0}}{{x}^{4}}+{}^{4}{{C}_{1}}{{x}^{3}}y+{}^{4}{{C}_{2}}{{x}^{2}}{{y}^{2}}+{}^{4}{{C}_{3}}x{{y}^{3}}+{}^{4}{{C}_{4}}{{y}^{4}}\]
Here \[y=1\]
Substituting \[y=1\] rewrite the equation
\[{{(x+1)}^{4}}={}^{4}{{C}_{0}}{{x}^{4}}+{}^{4}{{C}_{1}}{{x}^{3}}+{}^{4}{{C}_{2}}{{x}^{2}}+{}^{4}{{C}_{3}}x+{}^{4}{{C}_{4}}\]
Further binomial coefficients of each term has to be calculated
\[{}^{4}{{C}_{0}}=\dfrac{4!}{0!(4-0)!}\]
Further solving we get
\[{}^{4}{{C}_{0}}=1\]
Solve the next binomial coefficient
\[{}^{4}{{C}_{1}}=\dfrac{4!}{1!(4-1)!}\]
On simplifying we get
\[{}^{4}{{C}_{1}}=\dfrac{4!}{1!3!}\]
Further solving we get
\[{}^{4}{{C}_{1}}=4\]
Solving next binomial coefficient
\[{}^{4}{{C}_{2}}=\dfrac{4!}{2!(4-2)!}\]
On simplifying we get
\[{}^{4}{{C}_{2}}=\dfrac{4!}{2!2!}\]
Further solving we get
\[{}^{4}{{C}_{2}}=6\]
Solve the next binomial coefficient
\[{}^{4}{{C}_{3}}=\dfrac{4!}{3!(4-3)!}\]
On simplifying we get
\[{}^{4}{{C}_{3}}=\dfrac{4!}{3!1!}\]
Further solving we get
\[{}^{4}{{C}_{3}}=4\]
Solving the next binomial coefficient
\[{}^{4}{{C}_{4}}=\dfrac{4!}{4!(4-4)!}\]
On simplifying we get
\[{}^{4}{{C}_{4}}=\dfrac{4!}{4!0!}\]
Further solving we get
\[{}^{4}{{C}_{4}}=1\]
Now substituting all the calculated binomial coefficients in the binomial theorem formula and solve further to get the expansion
\[{{(x+1)}^{4}}={}^{4}{{C}_{0}}{{x}^{4}}+{}^{4}{{C}_{1}}{{x}^{3}}+{}^{4}{{C}_{2}}{{x}^{2}}+{}^{4}{{C}_{3}}x+{}^{4}{{C}_{4}}\]
On substituting we get expansion as
\[\therefore {{(x+1)}^{4}}={{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1\]
Note:The exponents of a binomial are always positive and can never be negative exponents. The degree of the binomial is the highest value of the exponent. The knowledge of combination and factorial is required to solve the expansion questions. A combination can be defined as each of the groups which can be formed by taking some or all of a number of objects. Always remember that \[{}^{n}{{C}_{n}}=1\] and \[{}^{n}{{C}_{0}}=1\]. Combination is used for unordered objects.
The Binomial Theorem is given by \[{{(x+y)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}y+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+...+{}^{n}{{C}_{n}}{{y}^{n}}\]
The equations that contain binomials are called binomial equations. Operations like addition, subtraction, multiplication, division, etc can be performed on Binomial expression.
Complete step by step answer:
The Binomial expansion of \[{{(x+1)}^{4}}\] according to the Binomial Theorem can be written as
\[{{(x+y)}^{4}}={}^{4}{{C}_{0}}{{x}^{4}}+{}^{4}{{C}_{1}}{{x}^{3}}y+{}^{4}{{C}_{2}}{{x}^{2}}{{y}^{2}}+{}^{4}{{C}_{3}}x{{y}^{3}}+{}^{4}{{C}_{4}}{{y}^{4}}\]
Here \[y=1\]
Substituting \[y=1\] rewrite the equation
\[{{(x+1)}^{4}}={}^{4}{{C}_{0}}{{x}^{4}}+{}^{4}{{C}_{1}}{{x}^{3}}+{}^{4}{{C}_{2}}{{x}^{2}}+{}^{4}{{C}_{3}}x+{}^{4}{{C}_{4}}\]
Further binomial coefficients of each term has to be calculated
\[{}^{4}{{C}_{0}}=\dfrac{4!}{0!(4-0)!}\]
Further solving we get
\[{}^{4}{{C}_{0}}=1\]
Solve the next binomial coefficient
\[{}^{4}{{C}_{1}}=\dfrac{4!}{1!(4-1)!}\]
On simplifying we get
\[{}^{4}{{C}_{1}}=\dfrac{4!}{1!3!}\]
Further solving we get
\[{}^{4}{{C}_{1}}=4\]
Solving next binomial coefficient
\[{}^{4}{{C}_{2}}=\dfrac{4!}{2!(4-2)!}\]
On simplifying we get
\[{}^{4}{{C}_{2}}=\dfrac{4!}{2!2!}\]
Further solving we get
\[{}^{4}{{C}_{2}}=6\]
Solve the next binomial coefficient
\[{}^{4}{{C}_{3}}=\dfrac{4!}{3!(4-3)!}\]
On simplifying we get
\[{}^{4}{{C}_{3}}=\dfrac{4!}{3!1!}\]
Further solving we get
\[{}^{4}{{C}_{3}}=4\]
Solving the next binomial coefficient
\[{}^{4}{{C}_{4}}=\dfrac{4!}{4!(4-4)!}\]
On simplifying we get
\[{}^{4}{{C}_{4}}=\dfrac{4!}{4!0!}\]
Further solving we get
\[{}^{4}{{C}_{4}}=1\]
Now substituting all the calculated binomial coefficients in the binomial theorem formula and solve further to get the expansion
\[{{(x+1)}^{4}}={}^{4}{{C}_{0}}{{x}^{4}}+{}^{4}{{C}_{1}}{{x}^{3}}+{}^{4}{{C}_{2}}{{x}^{2}}+{}^{4}{{C}_{3}}x+{}^{4}{{C}_{4}}\]
On substituting we get expansion as
\[\therefore {{(x+1)}^{4}}={{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1\]
Note:The exponents of a binomial are always positive and can never be negative exponents. The degree of the binomial is the highest value of the exponent. The knowledge of combination and factorial is required to solve the expansion questions. A combination can be defined as each of the groups which can be formed by taking some or all of a number of objects. Always remember that \[{}^{n}{{C}_{n}}=1\] and \[{}^{n}{{C}_{0}}=1\]. Combination is used for unordered objects.
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