Question

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

Answer 1

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$n$.${}^n{C_1}$