Question

How do you use the Binomial Theorem to expand \[{\left( {x + 1} \right)^4}?\]

Answer 1

Hint:
We have given a bi-quadratic binomial and we have to expand it with the help of the binomial theorem. We have to write the binomial expansion first of \[{\left( {a + bx} \right)^n}.\]Then we compare \[{\left( {x + 1} \right)^4}\]with the expansion and get the value of \[a\]and \[bx\] \[n.\]Then we apply the binomial expansion on this binomial then we find the value of \[{n_{{c_r}}}\]for each term by the formula of combination. This leads us to the required result.

Complete Step by step Solution:
We have given a bi-quadratic binomial expression and have to solve it by binomial theorem.
The bi-quadratic binomial is \[{\left( {x + 1} \right)^4}\]
By the binomial theorem we have \[{\left( {a + bx} \right)^n} = \sum\limits_{r = 0}^n {{n_{{c_r}}}{{\left( a \right)}^{n - r}}{{\left( {bx} \right)}^r}} \]
Comparing \[{\left( {x + 1} \right)^4}{\text{ with }}{\left( {a + bx} \right)^n}\]
We have \[a = 1,bx = x{\text{ and }}n = 4\]
So \[{(x + 1)^4} = \sum\limits_{r = 0}^4 {{4_{{c_r}}}{{\left( x \right)}^n}{{\left( 1 \right)}^{n - r}}} \]
Expanding the right-hand side we get
\[{(x + 1)^4} = {4_{{c_0}}}{\left( x \right)^1}{\left( 1 \right)^{4 - 0}} + {4_{{c_1}}}{\left( x \right)^1}{\left( 1 \right)^{4 - 1}} + {4_{{c_2}}}{\left( x \right)^2}{\left( 1 \right)^4} + {4_{{c_3}}}{\left( x \right)^1}{\left( 1 \right)^{4 - 1}} + {4_{{c_4}}}{\left( x \right)^1}{\left( 1 \right)^{4 - 4}}\]
\[ \Rightarrow {4_{{c_0}}}(1 \times 1) + {4_{{c_1}}}x + {4_{{c_2}}}{x^2} + {4_{{c_3}}}{x^3} + {4_{{c_4}}}{x^4}\]
As we know that \[{n_{{c_r}}} = \dfrac{{n!}}{{r!(n - r)!}}\]
So \[{4_{{c_0}}}\] is given as = \[{4_{{c_0}}} = \dfrac{{4!}}{{0!(4 - 0)!}}\] =\[\dfrac{{4!}}{{4!}}\]= \[1\]
So \[{4_{{c_1}}}\] is given as = \[{4_{{c_1}}} = \dfrac{{4!}}{{1!(4 - 1)!}} = \dfrac{{4!}}{{3!}} = \dfrac{{4 \times 3!}}{{3!}} = 4\]
So \[{4_{{c_2}}}\] is given as = \[{4_{{c_2}}} = \dfrac{{4!}}{{2!(4 - 2)!}} = \dfrac{{4!}}{{2! \times 2!}} = 6\]
So \[{4_{{c_3}}}\] is given as = \[{4_{{c_3}}} = \dfrac{{4!}}{{3!(4 - 3)!}} = \dfrac{{4!}}{{3!}} = 4\]
So \[{4_{{c_4}}}\] is given as \[{4_{{c_4}}} = \dfrac{{4!}}{{4!(4 - 4)!}} = \dfrac{{4!}}{{4! \times 0!}} = 1\]
\[{(x + 1)^4}\] is given as
\[ \Rightarrow {(x + 1)^4} = 1 + 4x + 6{x^2} + 4{x^3} + {x^4}\]
\[ \Rightarrow {(x + 1)^4} = {x^4} + 4{x^3} + 6{x^2} + 4x + 1\]
This is the expanded form of \[{(x + 1)^4}\]by binomial theorem.

Note:
Each of the different groups or sections which can be formed by taking the same or all of the numbers of objects irrespective of their arrangement is called the combination.
For example, the different combinations formed by three letters a,b,c. Thus only combinations formed of three letters taken at all ABC. Note that ab or ba are two different permutations but each gives the same combination.