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

seo-qna
Last updated date: 19th Jul 2024
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Answer
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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 $x$.