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Question: Prove that the expansion of \({{\left( 1-{{x}^{3}} \right)}^{n}}\) may be put into the form \({{\lef...

Prove that the expansion of (1−x3)n{{\left( 1-{{x}^{3}} \right)}^{n}} may be put into the form (1−x)3n+3nx(1−x)3n−2+3n(3n−3)1.2x2(1−x)3n−4+………{{\left( 1-x \right)}^{3n}}+3nx{{\left( 1-x \right)}^{3n-2}}+\dfrac{3n\left( 3n-3 \right)}{1.2}{{x}^{2}}{{\left( 1-x \right)}^{3n-4}}+\ldots \ldots \ldots .

Explanation

Solution

We will use only the first step of induction here. If this value becomes true then we can do the calculations better. We will also use the formula of binomial expansion is given by, (a+b)n=nC0anb0+nC1an−1b1+nC2an−2b2+……+nCna0bn{{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+\ldots \ldots +{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}} or by, (a+b)n=an+nan−1b1+n(n−1)2!an−2b2+……+bn{{\left( a+b \right)}^{n}}={{a}^{n}}+n{{a}^{n-1}}{{b}^{1}}+\dfrac{n\left( n-1 \right)}{2!}{{a}^{n-2}}{{b}^{2}}+\ldots \ldots +{{b}^{n}}. The formula of nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}, where n represents the total number of objects from which we have to choose and r represents the number of objects that are selected at a time.

Complete step-by-step answer :
Now, we need to prove the equation given by, (1−x3)n=(1−x)3n+3nx(1−x)3n−2+3n(3n−3)1.2x2(1−x)3n−4+………(i){{\left( 1-{{x}^{3}} \right)}^{n}}={{\left( 1-x \right)}^{3n}}+3nx{{\left( 1-x \right)}^{3n-2}}+\dfrac{3n\left( 3n-3 \right)}{1.2}{{x}^{2}}{{\left( 1-x \right)}^{3n-4}}+\ldots \ldots \ldots \left( i \right).
Now, we will substitute n = 1 in equation (i). This is because we are using the first step of induction here which is substitution as n = 1. If the equation (i) gets true for n = 1 then, we can proceed with the next step. But we will only use the first step here. This thing should be noticed. Thus, we get
(1−x3)1=(1−x)3(1)+3(1)x(1−x)3(1)−2+3(1)(3(1)−3)1.2x2(1−x)3(1)−4+…… ⇒1−x3=(1−x)3+3x(1−x)+0 ⇒1−x3=(1−x)3+3x(1−x) \begin{aligned} & {{\left( 1-{{x}^{3}} \right)}^{1}}={{\left( 1-x \right)}^{3\left( 1 \right)}}+3\left( 1 \right)x{{\left( 1-x \right)}^{3\left( 1 \right)-2}}+\dfrac{3\left( 1 \right)\left( 3\left( 1 \right)-3 \right)}{1.2}{{x}^{2}}{{\left( 1-x \right)}^{3\left( 1 \right)-4}}+\ldots \ldots \\\ & \Rightarrow 1-{{x}^{3}}={{\left( 1-x \right)}^{3}}+3x\left( 1-x \right)+0 \\\ & \Rightarrow 1-{{x}^{3}}={{\left( 1-x \right)}^{3}}+3x\left( 1-x \right) \\\ \end{aligned}
By taking the power n terms on both the sides of the expression, we get the term (1−x3)n=[(1−x)3+3x(1−x)]n...(iii){{\left( 1-{{x}^{3}} \right)}^{n}}={{\left[ {{\left( 1-x \right)}^{3}}+3x\left( 1-x \right) \right]}^{n}}...(iii)
Now, we will multiply and divide the right hand side of equation (iii) by (1−x)3{{\left( 1-x \right)}^{3}} and we get, [(1−x)3+3x(1−x)]n=[(1−x)3+3x(1−x)(1−x)3(1−x)3]n………(iv){{\left[ {{\left( 1-x \right)}^{3}}+3x\left( 1-x \right) \right]}^{n}}={{\left[ {{\left( 1-x \right)}^{3}}+3x\left( 1-x \right)\dfrac{{{\left( 1-x \right)}^{3}}}{{{\left( 1-x \right)}^{3}}} \right]}^{n}}\ldots \ldots \ldots \left( iv \right)
After this we will substitute the expression (iv) in equation (iii), we will get,
(1−x3)n=[(1−x)3(1−x)3[(1−x)3+3x(1−x)]]n ⇒(1−x3)n=[(1−x)3(1−x)3(1−x)3+3x(1−x)(1−x)3(1−x)3]n ⇒(1−x3)n=[(1−x)3+3x(1−x)(1−x)3(1−x)3]n \begin{aligned} & {{\left( 1-{{x}^{3}} \right)}^{n}}={{\left[ \dfrac{{{\left( 1-x \right)}^{3}}}{{{\left( 1-x \right)}^{3}}}\left[ {{\left( 1-x \right)}^{3}}+3x\left( 1-x \right) \right] \right]}^{n}} \\\ & \Rightarrow {{\left( 1-{{x}^{3}} \right)}^{n}}={{\left[ \dfrac{{{\left( 1-x \right)}^{3}}{{\left( 1-x \right)}^{3}}}{{{\left( 1-x \right)}^{3}}}+\dfrac{3x\left( 1-x \right){{\left( 1-x \right)}^{3}}}{{{\left( 1-x \right)}^{3}}} \right]}^{n}} \\\ & \Rightarrow {{\left( 1-{{x}^{3}} \right)}^{n}}={{\left[ {{\left( 1-x \right)}^{3}}+\dfrac{3x\left( 1-x \right){{\left( 1-x \right)}^{3}}}{{{\left( 1-x \right)}^{3}}} \right]}^{n}} \\\ \end{aligned}
Here we can see that the term (1−x)3{{\left( 1-x \right)}^{3}} is common in the right hand side. So, by taking out this term from the above equation we get,
(1−x3)n=[(1−x)3+(1+3x(1−x)(1−x)3)]n ⇒(1−x3)n=(1−x)3n(1+3x(1−x)(1−x)3)n ⇒(1−x3)n=(1−x)3n(1+3x(1−x)(1−x)−3)n \begin{aligned} & {{\left( 1-{{x}^{3}} \right)}^{n}}={{\left[ {{\left( 1-x \right)}^{3}}+\left( 1+\dfrac{3x\left( 1-x \right)}{{{\left( 1-x \right)}^{3}}} \right) \right]}^{n}} \\\ & \Rightarrow {{\left( 1-{{x}^{3}} \right)}^{n}}={{\left( 1-x \right)}^{3n}}{{\left( 1+\dfrac{3x\left( 1-x \right)}{{{\left( 1-x \right)}^{3}}} \right)}^{n}} \\\ & \Rightarrow {{\left( 1-{{x}^{3}} \right)}^{n}}={{\left( 1-x \right)}^{3n}}{{\left( 1+3x\left( 1-x \right){{\left( 1-x \right)}^{-3}} \right)}^{n}} \\\ \end{aligned}
We will now apply the formula, (a+b)x(a+b)y=(a+b)x+y{{\left( a+b \right)}^{x}}{{\left( a+b \right)}^{y}}={{\left( a+b \right)}^{x+y}} in the above expression. So, we get,
(1−x3)n=(1−x)3n[1+3x(1−x)1−3]n ⇒(1−x3)n=(1−x)3n[1+3x(1−x)−2]n………(v) \begin{aligned} & {{\left( 1-{{x}^{3}} \right)}^{n}}={{\left( 1-x \right)}^{3n}}{{\left[ 1+3x{{\left( 1-x \right)}^{1-3}} \right]}^{n}} \\\ & \Rightarrow {{\left( 1-{{x}^{3}} \right)}^{n}}={{\left( 1-x \right)}^{3n}}{{\left[ 1+3x{{\left( 1-x \right)}^{-2}} \right]}^{n}}\ldots \ldots \ldots \left( v \right) \\\ \end{aligned}
Now, we will apply the binomial expansion formula, (a+b)n=an+nan−1b1+n(n−1)2!an−2b2+……+bn{{\left( a+b \right)}^{n}}={{a}^{n}}+n{{a}^{n-1}}{{b}^{1}}+\dfrac{n\left( n-1 \right)}{2!}{{a}^{n-2}}{{b}^{2}}+\ldots \ldots +{{b}^{n}} in the above expression by substituting a =1 and b = 3x(1−x)−23x{{\left( 1-x \right)}^{-2}}. So, w get,
[1+3x(1−x)−2]n=1n+n(1)n−1(3x(1−x)−2)+n(n−1)2!(1)n−2(3x(1−x)−2)2+……+(3x(1−x)−2)n ⇒[1+3x(1−x)−2]n=1n+(3nx(1−x)−2)+n(n−1)2!9x2(1−x)−4+……+3nxn(1−x)−2n ⇒[1+3x(1−x)−2]n=1+3nx(1−x)−2+n(n−1)2!9x2(1−x)−4+……+3nxn(1−x)−2n………(vi) \begin{aligned} & {{\left[ 1+3x{{\left( 1-x \right)}^{-2}} \right]}^{n}}={{1}^{n}}+n{{\left( 1 \right)}^{n-1}}\left( 3x{{\left( 1-x \right)}^{-2}} \right)+\dfrac{n\left( n-1 \right)}{2!}{{\left( 1 \right)}^{n-2}}{{\left( 3x{{\left( 1-x \right)}^{-2}} \right)}^{2}}+\ldots \ldots +{{\left( 3x{{\left( 1-x \right)}^{-2}} \right)}^{n}} \\\ & \Rightarrow {{\left[ 1+3x{{\left( 1-x \right)}^{-2}} \right]}^{n}}={{1}^{n}}+\left( 3nx{{\left( 1-x \right)}^{-2}} \right)+\dfrac{n\left( n-1 \right)}{2!}9{{x}^{2}}{{\left( 1-x \right)}^{-4}}+\ldots \ldots +{{3}^{n}}{{x}^{n}}{{\left( 1-x \right)}^{-2n}} \\\ & \Rightarrow {{\left[ 1+3x{{\left( 1-x \right)}^{-2}} \right]}^{n}}=1+3nx{{\left( 1-x \right)}^{-2}}+\dfrac{n\left( n-1 \right)}{2!}9{{x}^{2}}{{\left( 1-x \right)}^{-4}}+\ldots \ldots +{{3}^{n}}{{x}^{n}}{{\left( 1-x \right)}^{-2n}}\ldots \ldots \ldots \left( vi \right) \\\ \end{aligned}
Now we will multiply equation (vi) by (1−x)3n{{\left( 1-x \right)}^{3n}} on both sides. So, we get,

& {{\left( 1-x \right)}^{3n}}{{\left[ 1+3x{{\left( 1-x \right)}^{-2}} \right]}^{n}}={{\left( 1-x \right)}^{3n}}\left[ 1+3nx{{\left( 1-x \right)}^{-2}}+\dfrac{n\left( n-1 \right)}{2!}9{{x}^{2}}{{\left( 1-x \right)}^{-4}}+\ldots +{{3}^{n}}{{x}^{n}}{{\left( 1-x \right)}^{-2n}} \right] \\\ & \Rightarrow {{\left( 1-x \right)}^{3n}}{{\left[ 1+3x{{\left( 1-x \right)}^{-2}} \right]}^{n}}={{\left( 1-x \right)}^{3n}}+3nx{{\left( 1-x \right)}^{3n}}{{\left( 1-x \right)}^{-2}}+\dfrac{n\left( n-1 \right)}{2!}9{{x}^{2}}{{\left( 1-x \right)}^{3n}}{{\left( 1-x \right)}^{-4}} \\\ & +\ldots +{{3}^{n}}{{x}^{n}}{{\left( 1-x \right)}^{3n}}{{\left( 1-x \right)}^{-2n}} \\\ \end{aligned}$$ We will consider the formula ${{\left( a+b \right)}^{x}}{{\left( a+b \right)}^{y}}={{\left( a+b \right)}^{x+y}}$ in the above expression. So, we get, $\begin{aligned} & {{\left( 1-x \right)}^{3n}}{{\left[ 1+3x{{\left( 1-x \right)}^{-2}} \right]}^{n}}={{\left( 1-x \right)}^{3n}}+3nx{{\left( 1-x \right)}^{3n-2}}+\dfrac{n\left( n-1 \right)}{2!}9{{x}^{2}}{{\left( 1-x \right)}^{3n-4}}+\ldots +{{3}^{n}}{{x}^{n}}{{\left( 1-x \right)}^{3n-2n}} \\\ & \Rightarrow {{\left( 1-x \right)}^{3n}}{{\left[ 1+3x{{\left( 1-x \right)}^{-2}} \right]}^{n}}={{\left( 1-x \right)}^{3n}}+3nx{{\left( 1-x \right)}^{3n-2}}+\dfrac{n\left( n-1 \right)}{2!}9{{x}^{2}}{{\left( 1-x \right)}^{3n-4}}+\ldots +{{3}^{n}}{{x}^{n}}{{\left( 1-x \right)}^{n}}\ldots \left( vii \right) \\\ \end{aligned}$ By looking at equation (iii), we know that, $1-{{x}^{3}}={{\left( 1-x \right)}^{3}}+3x\left( 1-x \right)$ and from equation (v), we know that, ${{\left( 1-{{x}^{3}} \right)}^{n}}={{\left( 1-x \right)}^{3n}}{{\left[ 1+3x{{\left( 1-x \right)}^{-2}} \right]}^{n}}$. So, combining equations (v) and (vii), we get, ${{\left( 1-{{x}^{3}} \right)}^{n}}={{\left( 1-x \right)}^{3n}}+3nx{{\left( 1-x \right)}^{3n-2}}+\dfrac{n\left( n-1 \right)}{2!}9{{x}^{2}}{{\left( 1-x \right)}^{3n-4}}+\ldots +{{3}^{n}}{{x}^{n}}{{\left( 1-x \right)}^{n}}$ Now, we will split the third term of the above expression, $9{{x}^{2}}$ into $3x\times 3x$ and write, $n\left( n-1 \right)9{{x}^{2}}$as $n\left( n-1 \right)3x\times 3x$. We can change the places of n by 3x and write it as $3xn\left( n-1 \right)3x$. Now we will multiply $\left( n-1 \right)$ by 3 and get, $3xn\left( n-1 \right)3x=3n\left( 3n-3 \right){{x}^{2}}$. We will substitute this in the above expression. So, we get, ${{\left( 1-{{x}^{3}} \right)}^{n}}={{\left( 1-x \right)}^{3n}}+3nx{{\left( 1-x \right)}^{3n-2}}+\dfrac{3n\left( 3n-3 \right)}{2!}{{x}^{2}}{{\left( 1-x \right)}^{3n-4}}+\ldots +{{3}^{n}}{{x}^{n}}{{\left( 1-x \right)}^{n}}$ In this question, we are supposed to prove the expression for finite terms. Thus we, get, ${{\left( 1-{{x}^{3}} \right)}^{n}}={{\left( 1-x \right)}^{3n}}+3nx{{\left( 1-x \right)}^{3n-2}}+\dfrac{3n\left( 3n-3 \right)}{2!}{{x}^{2}}{{\left( 1-x \right)}^{3n-4}}+\ldots $ As this is our required expression, we have proved the same. **Note** :We could have also used second and third steps of induction alternatively in this question. We have substituted n = 1, which is the same step as in induction. But after that we could have proceeded with the second step of induction but this will lead to a more complex solution. It can also take time in leading to the solution. The use of binomial expansion in this question is a must. This is because the expansion of ${{\left( 1-{{x}^{3}} \right)}^{n}}$ can only be done by this formula. As this solution is a bit lengthy, one should focus while doing calculations as in these types of lengthy questions one can make mistakes. As a result the answer gets wrong.