Question

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

Answer 1

Here we will apply the Binomial Expansion to solve the given problem.
The binomial theorem (or binomial expansion) describes the algebraic expansion of power of a binomial. According to the theorem, it is possible to expand the polynomial ${(x + y)^n}$.
First we have to Put the given value in the place of n and use the formula. Then solving this with the help of combination rule and factorial n and simplifying the result we will get the solution.

Formula used: Combination rule: $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Binomial expansion:
${(x + y)^n}{ = ^n}{C_0}{x^n}{ + ^n}{C_1}{x^{n - 1}}{y^1}{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......{ + ^n}{C_{n - 1}}x{y^{n - 1}}{ + ^n}{C_n}{y^n}$
Where $n \geqslant 0$, is an integer and each $^n{C_k}$ is a positive integer known as binomial coefficient.

Complete step-by-step solution:
We need to expand ${\left( {x + y} \right)^{10}}$.
Now we know that, according to binomial theorem it is possible to expand any nonnegative power of
$x + y$ into a sum of the form
${(x + y)^n}{ = ^n}{C_0}{x^n}{ + ^n}{C_1}{x^{n - 1}}{y^1}{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......{ + ^n}{C_{n - 1}}x{y^{n - 1}}{ + ^n}{C_n}{y^n}$, where $n \geqslant 0$, is an integer and each$^n{C_k}$ is a positive integer known as binomial coefficient.
Now we can use the binomial expansion putting $n = 10$ we get,
$\Rightarrow {(x + y)^{10}}{ = ^{10}}{C_0}{x^{10}}{y^0}{ + ^{10}}{C_1}{x^{10 - 1}}{y^1} + ...{ + ^{10}}{C_9}{x^{10 - 9}}{y^9}{ + ^{10}}{C_{10}}{x^0}{y^{10}}$
Solving we get,
$\Rightarrow {(x + y)^{10}}{ = ^{10}}{C_0}{x^{10}}{ + ^{10}}{C_1}{x^9}{y^1} + ...{ + ^{10}}{C_9}x{y^9}{ + ^{10}}{C_{10}}{y^{10}}$,
[Using,${x^0} = 1$].
Again, we can use the Combination rule,$^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$, we get,$\Rightarrow {(x + y)^{10}} = \dfrac{{10!}}{{0!\left( {10 - 0} \right)!}}{x^{10}} + \dfrac{{10!}}{{1!\left( {10 - 1} \right)!}}{x^9}{y^1} + .... + \dfrac{{10!}}{{9!\left( {10 - 9} \right)!}}x{y^9} + \dfrac{{10!}}{{10!\left( {10 - 10} \right)!}}{y^{10}}$
Solving we get,
$\Rightarrow {(x + y)^{10}} = \dfrac{{10!}}{{0!10!}}{x^{10}} + \dfrac{{10!}}{{1!9!}}{x^9}{y^1} + ... + \dfrac{{10!}}{{9!1!}}x{y^9} + \dfrac{{10!}}{{10!0!}}{y^{10}}$
Let us expanding the factorial term and we get
$\Rightarrow {(x + y)^{10}} = {x^{10}} + \dfrac{{10 \times 9!}}{{1 \times 9!}}{x^9}{y^1} + \dfrac{{10 \times 9 \times 8!}}{{2 \times 1 \times 8!}}{x^8}{y^2} + ... + \dfrac{{10 \times 9 \times 8!}}{{8! \times 2 \times 1}}{x^2}{y^8} + \dfrac{{10 \times 9!}}{{9! \times 1}}x{y^9} + {y^{10}}$
[Using the definition of factorial $n$, $n! = n(n - 1)(n - 2)(n - 4).......2.1$ and $0! = 1$]
Simplifying we get,
$\Rightarrow {(x + y)^{10}} = {x^{10}} + 10{x^9}{y^1} + \dfrac{{10 \times 9}}{2}{x^8}{y^2} + ... + \dfrac{{10 \times 9}}{{2 \times 1}}{x^2}{y^8} + 10x{y^9} + {y^{10}}$
Or,
$\Rightarrow {(x + y)^{10}} = {x^{10}} + 10{x^9}{y^1} + \left( {5 \times 9} \right){x^8}{y^2} + ... + \left( {5 \times 9} \right){x^2}{y^8} + 10x{y^9} + {y^{10}}$
Solving we get,
$\Rightarrow {(x + y)^{10}} = {x^{10}} + 10{x^9}{y^1} + 45{x^8}{y^2} + ... + 45{x^2}{y^8} + 10x{y^9} + {y^{10}}$

Hence expanding ${\left( {x + y} \right)^{10}}$ we get,
$\Rightarrow {(x + y)^{10}} = {x^{10}} + 10{x^9}{y^1} + 45{x^8}{y^2} + ... + 45{x^2}{y^8} + 10x{y^9} + {y^{10}}$

Note: In mathematics, a combination is a selection of items from a collection, such that the order of selection does not matter. For example, given three fruits, say an apple, an orange and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange.
A combination is the number of ways we can combine things, when the order does not matter.
For a combination,
$C\left( {n,r} \right){ = ^n}{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Where, factorial n is denoted by $n!$ and defined by
$n! = n(n - 1)(n - 2)(n - 4).......2.1$