Question

How do you expand\[{\left( {x + y} \right)^{10}}\]?

Answer 1

Hint: In this question we have to expand the given expression by using binomial formula which is given by\[{\left( {a + b} \right)^n} = {}^n{C_0}{a^n}{\left( b \right)^0} + {}^n{C_1}{a^{n - 1}}{\left( b \right)^1} + {}^n{C_2}{a^{n - 2}}{\left( b \right)^2} + {}^n{C_3}{a^{n - 3}}{\left( b \right)^3} + ....... + {}^n{C_n}{a^0}{\left( b \right)^n}\],
Now substituting the\[a\]and\[b\]value from the given expression and using the formula\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] we will get the required result.

Complete step-by-step answer:
Binomial expansion is given by the formula, \[{\left( {a + b} \right)^n} = {}^n{C_0}{a^n}{\left( b \right)^0} + {}^n{C_1}{a^{n - 1}}{\left( b \right)^1} + {}^n{C_2}{a^{n - 2}}{\left( b \right)^2} + {}^n{C_3}{a^{n - 3}}{\left( b \right)^3} + ....... + {}^n{C_n}{a^0}{\left( b \right)^n}\],
Where\[{}^n{C_0}\],\[{}^n{C_1}\],\[{}^n{C_2}\],\[{}^n{C_3}\]……..and\[{}^n{C_n}\]are the binomial coefficients .
So, now given expression is\[{\left( {x + y} \right)^{10}}\],
Now substituting\[a\]and\[b\]in the expansion where\[a = x\],\[b = y\]and \[n = 10\], then we get,
\[{\left( {x + y} \right)^{10}} = {}^{10}{C_0}{x^{10}}{\left( y \right)^0} + {}^{10}{C_1}{x^{10 - 1}}{\left( y \right)^1} + {}^{10}{C_2}{x^{10 - 2}}{\left( y \right)^2} + {}^{10}{C_3}{x^{10 - 3}}{\left( y \right)^3} + {}^{10}{C_4}{x^{10 - 4}}{\left( y \right)^4} + {}^{10}{C_5}{x^{10 - 5}}{\left( y \right)^5}\]\[ + {}^{10}{C_6}{x^{10 - 6}}{\left( y \right)^6} + {}^{10}{C_7}{x^{10 - 7}}{\left( y \right)^7} + {}^{10}{C_8}{x^{10 - 8}}{\left( y \right)^8} + {}^{10}{C_9}{x^{10 - 9}}{\left( y \right)^9} + {}^{10}{C_{10}}{x^{10 - 10}}{\left( y \right)^{10}}\],
Now simplifying the expansion using the formula\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], we get,
\[ \Rightarrow \]\[{\left( {x + y} \right)^{10}} = {x^{10}}{\left( y \right)^0} + 10{x^9}{\left( y \right)^1} + \dfrac{{10!}}{{2!\left( {10 - 8} \right)!}}{x^8}{\left( y \right)^2} + \dfrac{{10!}}{{3!\left( {10 - 3} \right)!}}{x^7}{\left( y \right)^3} + \dfrac{{10!}}{{4!\left( {10 - 4} \right)!}}{x^6}{\left( y \right)^4} + \]\[\dfrac{{10!}}{{5!\left( {10 - 5} \right)!}}{x^5}{\left( y \right)^5} + \dfrac{{10!}}{{6!\left( {10 - 6} \right)!}}{x^4}{\left( y \right)^6} + \dfrac{{10!}}{{7!\left( {10 - 7} \right)!}}{x^3}{\left( y \right)^7} + \dfrac{{10!}}{{8!\left( {10 - 8} \right)!}}{x^2}{\left( y \right)^8} + \dfrac{{10!}}{{9!\left( {10 - 9} \right)!}}{x^1}{\left( y \right)^9} + \]\[\dfrac{{10!}}{{10!\left( {10 - 10} \right)!}}{x^0}{\left( y \right)^{10}}\],
Now simplifying the expansion we get,
\[ \Rightarrow \]\[{\left( {x + y} \right)^{10}} = {x^{10}}{\left( y \right)^0} + 10{x^9}{\left( y \right)^1} + \dfrac{{10!}}{{2!\left( 2 \right)!}}{x^8}{\left( y \right)^2} + \dfrac{{10!}}{{3!\left( 7 \right)!}}{x^7}{\left( y \right)^3} + \dfrac{{10!}}{{4!\left( 6 \right)!}}{x^6}{\left( y \right)^4} + \dfrac{{10!}}{{5!\left( 5 \right)!}}{x^5}{\left( y \right)^5} + \]\[\dfrac{{10!}}{{6!\left( 4 \right)!}}{x^4}{\left( y \right)^6} + \dfrac{{10!}}{{7!\left( 3 \right)!}}{x^3}{\left( y \right)^7} + \dfrac{{10!}}{{8!\left( 2 \right)!}}{x^2}{\left( y \right)^8} + \dfrac{{10!}}{{9!\left( 1 \right)!}}{x^1}{\left( y \right)^9} + \dfrac{{10!}}{{10!\left( 0 \right)!}}{x^0}{\left( y \right)^{10}}\]
Now applying factorial formula i.e.,\[n! = n \times (n - 1) \times \left( {n - 2} \right) \times ........ \times 3 \times 2 \times 1\],we get,
\[ \Rightarrow {\left( {x + y} \right)^{10}} = {x^{10}}{\left( y \right)^0} + 10{x^9}{\left( y \right)^1} + \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {2 \times 1} \right)\left( {8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \right)}}{x^8}{\left( y \right)^2} + \]\[\dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {3 \times 2 \times 1} \right)\left( {7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \right)}}{x^7}{\left( y \right)^3} + \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {4 \times 3 \times 2 \times 1} \right)\left( {6 \times 5 \times 4 \times 3 \times 2 \times 1} \right)}}{x^6}{\left( y \right)^4} + \]\[\dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {5 \times 4 \times 3 \times 2 \times 1} \right)\left( {5 \times 4 \times 3 \times 2 \times 1} \right)}}{x^5}{\left( y \right)^5} + \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {6 \times 5 \times 4 \times 3 \times 2 \times 1} \right)\left( {4 \times 3 \times 2 \times 1} \right)}}{x^4}{\left( y \right)^6} + \]
\[\dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \right)\left( {3 \times 2 \times 1} \right)}}{x^3}{\left( y \right)^7} + \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \right)\left( {2 \times 1} \right)}}{x^2}{\left( y \right)^8} + \]
\[\dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \right)\left( 1 \right)}}{x^1}{\left( y \right)^9} + {x^0}{\left( y \right)^{10}}\].
Now simplifying by multiplying and dividing we get,
\[ \Rightarrow \]\[{\left( {x + y} \right)^{10}} = {x^{10}} + 10{x^9}y + 45{x^8}{y^2} + 120{x^7}{y^3} + 210{x^6}{y^4} + 252{x^5}{y^5} + 210{x^4}{y^6} + 120{x^3}{y^7} + 45{x^2}{y^8} + 10x{y^9} + {y^{10}}\]

Final Answer:
\[\therefore \]The expansion form of the given expression\[{\left( {x + y} \right)^{10}}\]is equal to\[{\left( {x + y} \right)^{10}} = {x^{10}} + 10{x^9}y + 45{x^8}{y^2} + 120{x^7}{y^3} + 210{x^6}{y^4} + 252{x^5}{y^5} + 210{x^4}{y^6} + 120{x^3}{y^7} + 45{x^2}{y^8} + 10x{y^9} + {y^{10}}\]

Note:
We use the binomial theorem to help us expand binomials to any given power without direct multiplication. As we have seen, multiplication can be time-consuming or even not possible in some cases.
Binomial expansion is given by the formula, \[{\left( {a + b} \right)^n} = {}^n{C_0}{a^n}{\left( b \right)^0} + {}^n{C_1}{a^{n - 1}}{\left( b \right)^1} + {}^n{C_2}{a^{n - 2}}{\left( b \right)^2} + {}^n{C_3}{a^{n - 3}}{\left( b \right)^3} + ....... + {}^n{C_n}{a^0}{\left( b \right)^n}\],
Where\[{}^n{C_0}\],,\[{}^n{C_2}\],\[{}^n{C_3}\]……..and\[{}^n{C_n}\]are the binomial coefficients .
According to the binomial theorem, the\[{\left( {r + 1} \right)^{th}}\]term in the expansion\[{\left( {a + b} \right)^n}\]is given by,
\[{T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}\], the term\[{\left( {r + 1} \right)^{th}}\]is the general term and the number of terms in the expansion\[{\left( {a + b} \right)^n}\]will be equal to\[\left( {n + 1} \right)\], Where\[{}^n{C_r}\]is the binomial coefficient, the sum of the binomial coefficients will be\[{2^n}\]because we know that,\[\sum\limits_{r = 0}^n {{}^n{C_r}} = {2^n}\], thus the sum of all odd binomial coefficients is equal to the sum of all even binomial coefficients and each is equal to\[{2^{n - 1}}\], the middle term depends upon the value of\[n\].\[{}^n{C_1}\]