Question

Show that the ${{n}^{th}}$ co-efficient in the expansion of ${{\left( 1-x \right)}^{-n}}$ is double of ${{\left( n-1 \right)}^{th}}$.

Answer 1

Hint: We start solving the problem by recalling the expansion of ${{\left( 1-x \right)}^{-n}}$. We then find the coefficient of the ${{\left( r+1 \right)}^{th}}$ term in the expansion. We use this and find the co-efficient of ${{n}^{th}}$ term in the expansion of ${{\left( 1-x \right)}^{-n}}$. We then find the co-efficient of ${{\left( n+1 \right)}^{th}}$ term in the expansion of ${{\left( 1-x \right)}^{-n}}$. We then take the ratio of obtained coefficients and make the necessary calculations to get the required result.

Complete step-by-step solution:
According to the problem, we need prove that the ${{n}^{th}}$ co-efficient in the expansion of ${{\left( 1-x \right)}^{-n}}$ is double of ${{\left( n-1 \right)}^{th}}$.
Let us first recall the expansion of ${{\left( 1-x \right)}^{-n}}$.
We know that the expansion of ${{\left( 1-x \right)}^{-n}}$ is defined as $1+nx+\dfrac{n\left( n+1 \right)}{2!}{{x}^{2}}+\dfrac{n\left( n+1 \right)\left( n+2 \right)}{3!}{{x}^{3}}+.......+\dfrac{n\left( n+1 \right)\left( n+2 \right)...\left( n+r-1 \right)}{r!}{{x}^{r}}+.....\infty $.
So, the co-efficient of ${{\left( r+1 \right)}^{th}}$ term in the expansion is defined as $\dfrac{n\left( n+1 \right)\left( n+2 \right).......\left( n+r-1 \right)}{r!}$.
Let us find the co-efficient of ${{n}^{th}}$ term.
So, we get ${{T}_{n}}=\dfrac{n\left( n+1 \right)\left( n+2 \right).......\left( n+n-1-1 \right)}{\left( n-1 \right)!}$.
$\Rightarrow {{T}_{n}}=\dfrac{n\left( n+1 \right)\left( n+2 \right).......\left( 2n-2 \right)}{\left( n-1 \right)!}$.
Let us multiply the terms in descending order.
$\Rightarrow {{T}_{n}}=\dfrac{\left( 2n-2 \right)\left( 2n-1 \right)\left( 2n \right).......n}{\left( n-1 \right)!}$.
Let us multiply $\left( n-1 \right)!$ in numerator and denominator.
$\Rightarrow {{T}_{n}}=\dfrac{\left( 2n-2 \right)\left( 2n-1 \right)\left( 2n \right).......n}{\left( n-1 \right)!}\times \dfrac{\left( n-1 \right)!}{\left( n-1 \right)!}$.
$\Rightarrow {{T}_{n}}=\dfrac{\left( 2n-2 \right)\left( 2n-1 \right)\left( 2n \right).......n}{\left( n-1 \right)!\left( n-1 \right)!}\times \left( n-1 \right)!$.
We know that $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
\[\Rightarrow {{T}_{n}}=\dfrac{\left( 2n-2 \right)\left( 2n-1 \right)\left( 2n \right).......\left( n \right)\left( n-1 \right)\left( n-2 \right)........2.1}{\left( n-1 \right)!\left( n-1 \right)!}\].
We can see that the numerator resembles $\left( 2n-2 \right)!$.
\[\Rightarrow {{T}_{n}}=\dfrac{\left( 2n-2 \right)!}{\left( n-1 \right)!\left( n-1 \right)!}\].
\[\Rightarrow {{T}_{n}}=\dfrac{\left( 2n-2 \right)!}{\left( n-1 \right)!\left( \left( 2n-2 \right)-\left( n-1 \right) \right)!}\].
We can see that this resembles with ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
\[\Rightarrow {{T}_{n}}={}^{2n-2}{{C}_{n-1}}\] ---(1).
Let us find the co-efficient of ${{\left( n-1 \right)}^{th}}$ term.
So, we get ${{T}_{n-1}}=\dfrac{\left( n \right)\left( n+1 \right)\left( n+2 \right).......\left( n+n-2-1 \right)}{\left( n-2 \right)!}$.
$\Rightarrow {{T}_{n-1}}=\dfrac{n\left( n+1 \right)\left( n+2 \right).......\left( 2n-3 \right)}{\left( n-2 \right)!}$.
Let us multiply the terms in descending order.
$\Rightarrow {{T}_{n-1}}=\dfrac{\left( 2n-3 \right)\left( 2n-2 \right)\left( 2n-1 \right).......n}{\left( n-2 \right)!}$.
Let us multiply $\left( n-1 \right)!$ in numerator and denominator.
$\Rightarrow {{T}_{n-1}}=\dfrac{\left( 2n-3 \right)\left( 2n-2 \right)\left( 2n-1 \right).......n}{\left( n-2 \right)!}\times \dfrac{\left( n-1 \right)!}{\left( n-1 \right)!}$.
$\Rightarrow {{T}_{n-1}}=\dfrac{\left( 2n-3 \right)\left( 2n-2 \right)\left( 2n-1 \right).......n}{\left( n-2 \right)!\left( n-1 \right)!}\times \left( n-1 \right)!$.
We know that $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
\[\Rightarrow {{T}_{n-1}}=\dfrac{\left( 2n-3 \right)\left( 2n-2 \right)\left( 2n-1 \right).......\left( n \right)\left( n-1 \right)\left( n-2 \right)........2.1}{\left( n-2 \right)!\left( n-1 \right)!}\].
We can see that the numerator resembles $\left( 2n-3 \right)!$.
\[\Rightarrow {{T}_{n-1}}=\dfrac{\left( 2n-3 \right)!}{\left( n-2 \right)!\left( n-1 \right)!}\].
\[\Rightarrow {{T}_{n-1}}=\dfrac{\left( 2n-3 \right)!}{\left( n-2 \right)!\left( \left( 2n-3 \right)-\left( n-2 \right) \right)!}\].
We can see that this resembles with ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
\[\Rightarrow {{T}_{n-1}}={}^{2n-3}{{C}_{n-2}}\]---(2).
Let us take the ratio of ${{T}_{n}}$ and ${{T}_{n-1}}$.
$\Rightarrow \dfrac{{{T}_{n}}}{{{T}_{n-1}}}=\dfrac{{}^{2n-2}{{C}_{n-1}}}{{}^{2n-3}{{C}_{n-2}}}$.
$\Rightarrow \dfrac{{{T}_{n}}}{{{T}_{n-1}}}=\dfrac{\dfrac{\left( 2n-2 \right)!}{\left( n-1 \right)!\left( n-1 \right)!}}{\dfrac{\left( 2n-3 \right)!}{\left( n-2 \right)!\left( n-1 \right)!}}$.
$\Rightarrow \dfrac{{{T}_{n}}}{{{T}_{n-1}}}=\dfrac{\left( 2n-2 \right)}{\left( n-1 \right)}$.
$\Rightarrow \dfrac{{{T}_{n}}}{{{T}_{n-1}}}=2$.
$\Rightarrow {{T}_{n}}=2\times {{T}_{n-1}}$.
$\therefore$ We have proved that the co-efficient of ${{n}^{th}}$ co-efficient in the expansion of ${{\left( 1-x \right)}^{-n}}$ is double of ${{\left( n-1 \right)}^{th}}$.

Note: We should not take the binomial expansion ${{\left( 1-x \right)}^{-n}}$ as $\dfrac{1}{{{\left( 1-x \right)}^{n}}}$ and expand it in the denominator. We get infinity terms in the binomial expansion of the negative powers of the given terms. We can also verify this result also by taking the values for the variable n. We can also expect problems with expansion of rational powers instead of negative powers that we had in the problem.