
How do you use the binomial theorem to expand ${\left( {x + y} \right)^6}$ ?
Answer
459k+ views
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.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 12 Economics: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What is a transformer Explain the principle construction class 12 physics CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?
