Question

How do you find the binomial expansion of \[{\left( {x + y} \right)^7}\]?

Answer 1

Hint:
Given a binomial expression with some exponent. We have to find the binomial expansion of the given expression. We will apply the binomial formula \[{\left( {a + b} \right)^n} = {}^n{C_0}{a^n}{\left( b \right)^0} + {}^n{C_1}{a^{n - 1}}{\left( b \right)^1} + {}^n{C_2}{a^{n - 2}}{\left( b \right)^2} + {}^n{C_3}{a^{n - 3}}{\left( b \right)^3} + \ldots \ldots + {}^n{C_n}{a^0}{\left( b \right)^n}\]. Then, we will substitute the value of a, b and n from the given expression, we will simplify the expression.

Formula used:
The binomial expansion formula is given by:
\[{\left( {a + b} \right)^n} = {}^n{C_0}{a^n}{\left( b \right)^0} + {}^n{C_1}{a^{n - 1}}{\left( b \right)^1} + {}^n{C_2}{a^{n - 2}}{\left( b \right)^2} + {}^n{C_3}{a^{n - 3}}{\left( b \right)^3} + \ldots \ldots + {}^n{C_n}{a^0}{\left( b \right)^n}\]
Where\[{}^n{C_0}\], \[{}^n{C_1}\], \[{}^n{C_2}\], …. and \[{}^n{C_n}\] are called binomial coefficients.
The formula for \[{}^n{C_r}\] is given by:
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

Complete step by step solution:
We are given the binomial expression, \[{\left( {x + y} \right)^7}\]. First, we will apply the binomial formula by substituting \[a = x\], \[b = y\] and \[n = 7\].
\[{\left( {x + y} \right)^7} = {}^7{C_0}{x^7}{\left( y \right)^0} + {}^7{C_2}{x^{7 - 1}}{\left( y \right)^1} + {}^7{C_2}{x^{7 - 2}}{\left( y \right)^2} + {}^7{C_3}{x^{7 - 3}}{\left( y \right)^3} + {}^7{C_4}{x^{7 - 4}}{\left( y \right)^4} + {}^7{C_5}{x^{7 - 5}}{\left( y \right)^5} + {}^7{C_6}{x^{7 - 6}}{\left( y \right)^6} + {}^7{C_7}{x^{7 - 7}}{\left( y \right)^7}\]
On simplifying the expression, we get:
\[ \Rightarrow {\left( {x + y} \right)^7} = {}^7{C_0}{x^7}\left( 1 \right) + {}^7{C_2}{x^6}y + {}^7{C_2}{x^5}{\left( y \right)^2} + {}^7{C_3}{x^4}{\left( y \right)^3} + {}^7{C_4}{x^3}{\left( y \right)^4} + {}^7{C_5}{x^2}{\left( y \right)^5} + {}^7{C_6}x{\left( y \right)^6} + {}^7{C_7}\left( 1 \right){\left( y \right)^7}\]
Now, we will apply the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] to the expression.
\[ \Rightarrow {\left( {x + y} \right)^7} = \dfrac{{7!}}{{0!\left( {7 - 0} \right)!}}{x^7} + \dfrac{{7!}}{{1!\left( {7 - 1} \right)!}}{x^6}y + \dfrac{{7!}}{{2!\left( {7 - 2} \right)!}}{x^5}{\left( y \right)^2} + \dfrac{{7!}}{{3!\left( {7 - 3} \right)!}}{x^4}{\left( y \right)^3} + \dfrac{{7!}}{{4!\left( {7 - 4} \right)!}}{x^3}{\left( y \right)^4} + \dfrac{{7!}}{{5!\left( {7 - 5} \right)!}}{x^2}{\left( y \right)^5} + \dfrac{{7!}}{{6!\left( {7 - 6} \right)!}}x{\left( y \right)^6} + \dfrac{{7!}}{{7!\left( {7 - 7} \right)!}}\left( 1 \right){\left( y \right)^7}\]
Now, on simplifying the expansion by applying the factorial formula \[n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times \ldots \ldots \times 3 \times 2 \times 1\], we get:
\[ \Rightarrow {\left( {x + y} \right)^7} = \left( 1 \right){x^7} + \dfrac{{7 \times 6!}}{{1!6!}}{x^6}y + \dfrac{{7 \times 6 \times 5!}}{{2!5!}}{x^5}{\left( y \right)^2} + \dfrac{{7 \times 6 \times 5 \times 4!}}{{3 \times 2 \times 1 \times 4!}}{x^4}{\left( y \right)^3} + \dfrac{{7 \times 6 \times 5 \times 4!}}{{3 \times 2 \times 1 \times 4!}}{x^3}{\left( y \right)^4} + \dfrac{{7 \times 6 \times 5!}}{{5! \times 2 \times 1}}{x^2}{\left( y \right)^5} + \dfrac{{7 \times 6!}}{{6!1!}}x{\left( y \right)^6} + \left( 1 \right)\left( 1 \right){\left( y \right)^7}\]
Now, cancel out the common terms, we get:
\[ \Rightarrow {\left( {x + y} \right)^7} = {x^7} + 7{x^6}y + 21{x^5}{y^2} + 35{x^4}{y^3} + 35{x^3}{y^4} + 21{x^2}{y^5} + 7x{y^6} + {y^7}\]

Final answer: Hence, the binomial expansion of \[{\left( {x + y} \right)^7}\] is \[{x^7} + 7{x^6}y + 21{x^5}{y^2} + 35{x^4}{y^3} + 35{x^3}{y^4} + 21{x^2}{y^5} + 7x{y^6} + {y^7}\]

Note:
The students must note that there is an alternate method to solve such types of questions. In the alternate method there is a need to know the relation between binomial expansion and Pascal’s triangle.
In Pascal's triangle, we can write the expansion of \[{\left( {a + b} \right)^n}\] from \[n + 1\] row number. The terms without coefficients of \[{\left( {a + b} \right)^n}\] in Pascal’s triangle are \[{a^n},{a^{n - 1}}b,{a^{n - 2}}{b^2},{a^{n - 3}}{b^3}, \ldots ,{b^n}\]
Here, \[a = x\], \[b = y\] and \[n = 7\]
The eighth row of the Pascal triangle is given by:
1 7 21 35 35 21 7 1 …… (1)
Also, the terms without the coefficients of \[{\left( {a + b} \right)^7}\]
\[{\left( {a + b} \right)^7} = {a^7} + {a^6}b + {a^5}{b^2} + {a^4}{b^3} + {a^3}{b^4} + {a^2}{b^5} + a{b^6} + {b^7}\] …… (2)
Now, we will combine the equation (1) and (2), to write the expansion:
\[ \Rightarrow {\left( {a + b} \right)^7} = {a^7} + 7{a^6}b + 21{a^5}{b^2} + 35{a^4}{b^3} + 35{a^3}{b^4} + 21{a^2}{b^5} + 7a{b^6} + {b^7}\]
Now, substitute \[a = x\] and \[b = y\] into the expression.
\[ \Rightarrow {\left( {x + y} \right)^7} = {x^7} + 7{x^6}y + 21{x^5}{y^2} + 35{x^4}{y^3} + 35{x^3}{y^4} + 21{x^2}{y^5} + 7x{y^6} + {y^7}\]
Therefore, the required expansion is \[{x^7} + 7{x^6}y + 21{x^5}{y^2} + 35{x^4}{y^3} + 35{x^3}{y^4} + 21{x^2}{y^5} + 7x{y^6} + {y^7}\]