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

Expand the following binomial: (1+x2)7{\left( {1 + \dfrac{x}{2}} \right)^7}

Explanation

Solution

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

(1+x)n=nC0(1)n(x)0+nC1(1)n−1(x)1+nC2(1)n−2(x)2+.....+nCn−1(1)1(x)n−1+nCn(1)0(x)n ⇒(1+x)n=nC0+nC1(x)+nC2(x)2+.....+nCn−1(x)n−1+nCn(x)n  {\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 nCr=n!r!(n−r)!  →(1){}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\; \to {\text{(1)}}

Here for the binomial expansion of (1+x2)7{\left( {1 + \dfrac{x}{2}} \right)^7}, xx is replaced by x2\dfrac{x}{2} and the value of nn is 7.

(1+x2)7=7C0+7C1(x2)+7C2(x2)2+7C3(x2)3+7C4(x2)4+7C5(x2)5+7C6(x2)6+7C7(x2)7 →(2){\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

7C0=7!0!(7−0)!  =7!0!7!  =1 [∵0!=1],7C1=7!1!(7−1)!  =7.6!6!  =7 7C2=7!2!(7−2)!  =7.6.5!2.1.5!  =7×62=21,7C3=7!3!(7−3)!  =7.6.5.4!3.2.1.4!  =7×6×53×2=35 7C4=7.6.5.4!4!(7−4)!  =7.6.5.4!4!3.2.1!  =7C3=35,7C5=7.6.5!5!(7−5)!  =7.6.5!5!2.1!  =7C2=21 7C6=7.6!6!(7−6)!  =7.6!6!1!  =7C1=7,7C7=7!7!(7−7)!  =7!7!0!  =7C0=1   {}^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
(1+x2)7=1+7x2+21(x)24+35(x)38+35(x)416+21(x)532+7(x)664+(x)7128{\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 (1+x2)7{\left( {1 + \dfrac{x}{2}} \right)^7}.

Note- The general form of binomial expansion is (x+y)n=nC0(x)n(y)0+nC1(x)n−1(y)1+nC2(x)n−2(y)2+.....+nCn−1(x)1(y)n−1+nCn(x)0(y)n{\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 xx by 1 and yy by xx.