Courses
Courses for Kids
Free study material
Free LIVE classes
More
Questions & Answers
seo-qna
LIVE
Join Vedantu’s FREE Mastercalss

Expand the following binomial: ${\left( {1 + \dfrac{x}{2}} \right)^7}$

Answer
VerifiedVerified
363.9k+ views
Hint- Here, we will proceed by using one of the special forms of the general form of binomial expansion.
As we know that according to special form of binomial theorem of expansion, we have
\[
  {\left( {1 + x} \right)^n} = {}^n{C_0}{\left( 1 \right)^n}{\left( x \right)^0} + {}^n{C_1}{\left( 1 \right)^{n - 1}}{\left( x \right)^1} + {}^n{C_2}{\left( 1 \right)^{n - 2}}{\left( x \right)^2} + ..... + {}^n{C_{n - 1}}{\left( 1 \right)^1}{\left( x \right)^{n - 1}} + {}^n{C_n}{\left( 1 \right)^0}{\left( x \right)^n} \\
   \Rightarrow {\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}\left( x \right) + {}^n{C_2}{\left( x \right)^2} + ..... + {}^n{C_{n - 1}}{\left( x \right)^{n - 1}} + {}^n{C_n}{\left( x \right)^n} \\
 \]
where ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\; \to {\text{(1)}}$

Here for the binomial expansion of ${\left( {1 + \dfrac{x}{2}} \right)^7}$, $x$ is replaced by $\dfrac{x}{2}$ and the value of $n$ is 7.

${\left( {1 + \dfrac{x}{2}} \right)^7} = {}^7{C_0} + {}^7{C_1}\left( {\dfrac{x}{2}} \right) + {}^7{C_2}{\left( {\dfrac{x}{2}} \right)^2} + {}^7{C_3}{\left( {\dfrac{x}{2}} \right)^3} + {}^7{C_4}{\left( {\dfrac{x}{2}} \right)^4} + {}^7{C_5}{\left( {\dfrac{x}{2}} \right)^5} + {}^7{C_6}{\left( {\dfrac{x}{2}} \right)^6} + {}^7{C_7}{\left( {\dfrac{x}{2}} \right)^7}{\text{ }} \to {\text{(2)}}$

Now using equation (1), we can write
\[
  {}^7{C_0} = \dfrac{{7!}}{{0!\left( {7 - 0} \right)!}}\; = \dfrac{{7!}}{{0!7!}}\; = 1{\text{ }}\left[ {\because 0! = 1} \right],{}^7{C_1} = \dfrac{{7!}}{{1!\left( {7 - 1} \right)!}}\; = \dfrac{{7.6!}}{{6!}}\; = 7 \\
  {}^7{C_2} = \dfrac{{7!}}{{2!\left( {7 - 2} \right)!}}\; = \dfrac{{7.6.5!}}{{2.1.5!}}\; = \dfrac{{7 \times 6}}{2} = 21,{}^7{C_3} = \dfrac{{7!}}{{3!\left( {7 - 3} \right)!}}\; = \dfrac{{7.6.5.4!}}{{3.2.1.4!}}\; = \dfrac{{7 \times 6 \times 5}}{{3 \times 2}} = 35 \\
  {}^7{C_4} = \dfrac{{7.6.5.4!}}{{4!\left( {7 - 4} \right)!}}\; = \dfrac{{7.6.5.4!}}{{4!3.2.1!}}\; = {}^7{C_3} = 35,{}^7{C_5} = \dfrac{{7.6.5!}}{{5!\left( {7 - 5} \right)!}}\; = \dfrac{{7.6.5!}}{{5!2.1!}}\; = {}^7{C_2} = 21 \\
  {}^7{C_6} = \dfrac{{7.6!}}{{6!\left( {7 - 6} \right)!}}\; = \dfrac{{7.6!}}{{6!1!}}\; = {}^7{C_1} = 7,{}^7{C_7} = \dfrac{{7!}}{{7!\left( {7 - 7} \right)!}}\; = \dfrac{{7!}}{{7!0!}}\; = {}^7{C_0} = 1 \\
    \\
 \]
Now substituting all the above calculated values in equation (2), we get
${\left( {1 + \dfrac{x}{2}} \right)^7} = 1 + \dfrac{{7x}}{2} + \dfrac{{21{{\left( x \right)}^2}}}{4} + \dfrac{{35{{\left( x \right)}^3}}}{8} + \dfrac{{35{{\left( x \right)}^4}}}{{16}} + \dfrac{{21{{\left( x \right)}^5}}}{{32}} + \dfrac{{7{{\left( x \right)}^6}}}{{64}} + \dfrac{{{{\left( x \right)}^7}}}{{128}}$
The above equation shows the binomial expansion for ${\left( {1 + \dfrac{x}{2}} \right)^7}$.

Note- The general form of binomial expansion is \[{\left( {x + y} \right)^n} = {}^n{C_0}{\left( x \right)^n}{\left( y \right)^0} + {}^n{C_1}{\left( x \right)^{n - 1}}{\left( y \right)^1} + {}^n{C_2}{\left( x \right)^{n - 2}}{\left( y \right)^2} + ..... + {}^n{C_{n - 1}}{\left( x \right)^1}{\left( y \right)^{n - 1}} + {}^n{C_n}{\left( x \right)^0}{\left( y \right)^n}\] and in this problem, its special form is used by replacing $x$ by 1 and $y$ by $x$.

Last updated date: 24th Sep 2023
Total views: 363.9k
Views today: 8.63k