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Question
AnswersHow do you use the binomial theorem to expand ${\left( {x + y} \right)^6}$ ?
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Hint: In this question, we are given an expression and we have been asked to expand the expression using binomial theorem. Please ensure that you are aware about the binomial theorem formula. Start putting the value of $r$, starting from $0$ and increasing till the value of $n$. Then, simplify all the terms and add them to get a final answer. Also, ensure that you are aware about the formula of combinations.
Formula used: 1) Binomial Theorem - ${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}{y^r}} $
2) Combinations formula - $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}$
Complete step-by-step solution:
Let us start putting the values of $r$ from $0$ in the formula${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}{y^r}} $.$ \Rightarrow {\left( {x + y} \right)^6} = \sum\limits_{r = 0}^6 {^n{C_r}{x^{n - r}}{y^r}} $
We will first put $r = 0$ then $r = 1$ and take it till $r = 6$. Then, we will add all the terms (As formula requires adding the terms.)
$ \Rightarrow {\left( {x + y} \right)^6}{ = ^6}{C_0}{x^{6 - 0}}{y^0}{ + ^6}{C_1}{x^{6 - 1}}{y^1}{ + ^6}{C_2}{x^{6 - 2}}{y^2}{ + ^6}{C_3}{x^{6 - 3}}{y^3}{ + ^6}{C_4}{x^{6 - 4}}{y^4}{ + ^6}{C_5}{x^{6 - 5}}{y^5}{ + ^6}{C_6}{x^{6 - 6}}{y^6}$Now, we know that if $r + p = n$ then,$^n{C_r}{ = ^n}{C_p}$
Using this identity, we can say that $^6{C_0}{ = ^6}{C_6}$, $^6{C_1}{ = ^6}{C_5}$ and$^6{C_2}{ = ^6}{C_4}$.
Thus, we only have to find the value of $^6{C_0}$, $^6{C_1}$, $^6{C_2}$ and $^6{C_3}$.
Let us first find their values individually.
${ \Rightarrow ^6}{C_0} = \dfrac{{6!}}{{0! \times \left( {6 - 0} \right)!}}$$ = \dfrac{{6!}}{{1 \times 6!}} = 1$ …. $(0! = 1)$
Hence, $^6{C_0}{ = ^6}{C_6} = 1$
${ \Rightarrow ^6}{C_1} = \dfrac{{6!}}{{1! \times \left( {6 - 1} \right)!}} = \dfrac{{6!}}{{5!}} = \dfrac{{6 \times 5!}}{{5!}} = 6$
Hence, $^6{C_1}{ = ^6}{C_5} = 6$
${ \Rightarrow ^6}{C_2} = \dfrac{{6!}}{{\left( {6 - 2} \right)! \times 2!}} = \dfrac{{6!}}{{4! \times 2!}} = \dfrac{{6 \times 5 \times 4!}}{{4! \times 2}} = \dfrac{{30}}{2} = 15$
Hence, $^6{C_2}{ = ^6}{C_4} = 15$
${ \Rightarrow ^6}{C_3} = \dfrac{{6!}}{{3! \times \left( {6 - 3} \right)!}} = \dfrac{{6!}}{{3! \times 3!}} = \dfrac{{6 \times 5 \times 4 \times 3!}}{{3! \times 3 \times 2}} = 20$
Now, we will put these values in the expanded expression.
$ \Rightarrow {\left( {x + y} \right)^6} = {x^6} + 6{x^5}y + 15{x^4}{y^2} + 20{x^3}{y^3} + 15{x^2}{y^4} + 6x{y^5} + {y^6}$
Now, since the powers of every term are different, we cannot simplify them further.
Hence, our final answer is $ {x^6} + 6{x^5}y + 15{x^4}{y^2} + 20{x^3}{y^3} + 15{x^2}{y^4} + 6x{y^5} + {y^6}$
Note: The two letters in the word – binomial, “bi” means two. A binomial has only two terms. But, when these are multiplied with it, what do we get? To get an answer to this question, we use binomial theorem.
You need not memorize the formula. A trick to write the answer is that give the highest power to x and lowest to y. Then, gradually, with the increase in terms, reduce the powers of x by 1 and increase the powers of y by 1. Continue this till the power of x reaches 0 and power of y reaches maximum.
Formula used: 1) Binomial Theorem - ${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}{y^r}} $
2) Combinations formula - $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}$
Complete step-by-step solution:
Let us start putting the values of $r$ from $0$ in the formula${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}{y^r}} $.$ \Rightarrow {\left( {x + y} \right)^6} = \sum\limits_{r = 0}^6 {^n{C_r}{x^{n - r}}{y^r}} $
We will first put $r = 0$ then $r = 1$ and take it till $r = 6$. Then, we will add all the terms (As formula requires adding the terms.)
$ \Rightarrow {\left( {x + y} \right)^6}{ = ^6}{C_0}{x^{6 - 0}}{y^0}{ + ^6}{C_1}{x^{6 - 1}}{y^1}{ + ^6}{C_2}{x^{6 - 2}}{y^2}{ + ^6}{C_3}{x^{6 - 3}}{y^3}{ + ^6}{C_4}{x^{6 - 4}}{y^4}{ + ^6}{C_5}{x^{6 - 5}}{y^5}{ + ^6}{C_6}{x^{6 - 6}}{y^6}$Now, we know that if $r + p = n$ then,$^n{C_r}{ = ^n}{C_p}$
Using this identity, we can say that $^6{C_0}{ = ^6}{C_6}$, $^6{C_1}{ = ^6}{C_5}$ and$^6{C_2}{ = ^6}{C_4}$.
Thus, we only have to find the value of $^6{C_0}$, $^6{C_1}$, $^6{C_2}$ and $^6{C_3}$.
Let us first find their values individually.
${ \Rightarrow ^6}{C_0} = \dfrac{{6!}}{{0! \times \left( {6 - 0} \right)!}}$$ = \dfrac{{6!}}{{1 \times 6!}} = 1$ …. $(0! = 1)$
Hence, $^6{C_0}{ = ^6}{C_6} = 1$
${ \Rightarrow ^6}{C_1} = \dfrac{{6!}}{{1! \times \left( {6 - 1} \right)!}} = \dfrac{{6!}}{{5!}} = \dfrac{{6 \times 5!}}{{5!}} = 6$
Hence, $^6{C_1}{ = ^6}{C_5} = 6$
${ \Rightarrow ^6}{C_2} = \dfrac{{6!}}{{\left( {6 - 2} \right)! \times 2!}} = \dfrac{{6!}}{{4! \times 2!}} = \dfrac{{6 \times 5 \times 4!}}{{4! \times 2}} = \dfrac{{30}}{2} = 15$
Hence, $^6{C_2}{ = ^6}{C_4} = 15$
${ \Rightarrow ^6}{C_3} = \dfrac{{6!}}{{3! \times \left( {6 - 3} \right)!}} = \dfrac{{6!}}{{3! \times 3!}} = \dfrac{{6 \times 5 \times 4 \times 3!}}{{3! \times 3 \times 2}} = 20$
Now, we will put these values in the expanded expression.
$ \Rightarrow {\left( {x + y} \right)^6} = {x^6} + 6{x^5}y + 15{x^4}{y^2} + 20{x^3}{y^3} + 15{x^2}{y^4} + 6x{y^5} + {y^6}$
Now, since the powers of every term are different, we cannot simplify them further.
Hence, our final answer is $ {x^6} + 6{x^5}y + 15{x^4}{y^2} + 20{x^3}{y^3} + 15{x^2}{y^4} + 6x{y^5} + {y^6}$
Note: The two letters in the word – binomial, “bi” means two. A binomial has only two terms. But, when these are multiplied with it, what do we get? To get an answer to this question, we use binomial theorem.
You need not memorize the formula. A trick to write the answer is that give the highest power to x and lowest to y. Then, gradually, with the increase in terms, reduce the powers of x by 1 and increase the powers of y by 1. Continue this till the power of x reaches 0 and power of y reaches maximum.