Question

Prove that ${}^{m}{{C}_{1}}{}^{n}{{C}_{m}}-{}^{m}{{C}_{2}}{}^{2n}{{C}_{m}}+{}^{m}{{C}_{3}}{}^{3n}{{C}_{m}}..............\,=\,{{(-1)}^{m-1}}{{n}^{m}}$

Answer 1

Hint: ${}^{n}{{C}_{r}}$ denotes the coefficient of ${{x}^{r}}$ in the binomial expansion of ${{(1+x)}^{n}}$. Consider ${}^{kn}{{C}_{m}}$ in the expansion as coefficients of terms of ${{(1+x)}^{n}}$and substitute their values. Use the following binomial expansions,
${{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{0}}x+{}^{n}{{C}_{0}}{{x}^{2}}+\,......\,+{}^{n}{{C}_{n}}{{x}^{n}}$
${{(1-x)}^{n}}={}^{n}{{C}_{0}}-{}^{n}{{C}_{0}}x+{}^{n}{{C}_{0}}{{x}^{2}}-\,......\,+(-1)n{}^{n}{{C}_{n}}{{x}^{n}}$

Complete step by step answer:
We know that ${}^{n}{{C}_{r}}$ denotes the coefficient of ${{x}^{r}}$ in the binomial expansion of ${{(1+x)}^{n}}$, where r arbitrary constant that can take values from 0 to n.
By this logic we can deduce the following,
${}^{n}{{C}_{m}}$ is the coefficient of ${{x}^{m}}$ in the expansion of ${{(1+x)}^{n}}$
${}^{2n}{{C}_{m}}$ is the coefficient of ${{x}^{m}}$ in the expansion of ${{(1+x)}^{2n}}$
${}^{3n}{{C}_{m}}$ is the coefficient of ${{x}^{m}}$ in the expansion of ${{(1+x)}^{3n}}$and so on.
LHS $={}^{m}{{C}_{1}}{}^{n}{{C}_{m}}-{}^{m}{{C}_{2}}{}^{2n}{{C}_{m}}+{}^{m}{{C}_{3}}{}^{3n}{{C}_{m}}-....$
Substituting this, we can rewrite LHS,
 $LHS={}^{m}{{C}_{1}}\times $(coefficient of ${{x}^{m}}$ in ${{(1+x)}^{n}}$) - ${}^{m}{{C}_{2}}\times $(coefficient of ${{x}^{m}}$ in ${{(1+x)}^{2n}}$) + ${}^{m}{{C}_{3}}\times $(coefficient of ${{x}^{m}}$ in ${{(1+x)}^{3n}}$) …………
Now collecting all the coefficients of ${{x}^{m}}$ together from all the terms of the above expansion, we get,
$LHS=$ Coefficient of ${{x}^{m}}$ in \[\left\{ {}^{m}{{C}_{1}}{{(1+x)}^{n}}-{}^{m}{{C}_{2}}{{(1+x)}^{2n}}+\,..... \right\}\]
Adding and subtracting ${}^{m}{{C}_{0}}$ to the expansion,
$LHS=$ Coefficient of ${{x}^{m}}$ in $\left\{ {}^{m}{{C}_{0}}-{}^{m}{{C}_{0}}+{}^{m}{{C}_{1}}{{(1+x)}^{n}}-{}^{m}{{C}_{2}}{{(1+x)}^{2n}}+\,.... \right\}$
We know that the value of ${}^{m}{{C}_{0}}$ is 1. Substituting this value to the first ${}^{m}{{C}_{0}}$,
$LHS=$ Coefficient of ${{x}^{m}}$ in $\left\{ 1-{}^{m}{{C}_{0}}+{}^{m}{{C}_{1}}{{(1+x)}^{n}}-{}^{m}{{C}_{2}}{{(1+x)}^{2n}}+\,.... \right\}$
Taking -1 common from all the terms inside of the expansion other than 1,
$LHS=$ Coefficient of ${{x}^{m}}$ in $\left\{ 1-({}^{m}{{C}_{0}}-{}^{m}{{C}_{1}}{{(1+x)}^{n}}+{}^{m}{{C}_{2}}{{(1+x)}^{2n}}-\,.....) \right\}...........(1)$
We know the following binomial expansion,
${{(1-y)}^{m}}={}^{m}{{C}_{0}}-{}^{m}{{C}_{1}}y+{}^{m}{{C}_{2}}{{y}^{2}}-\,.........\,$
Substitute $y={{(1+x)}^{n}}$, we get,
${{\left[ 1-{{(1+x)}^{n}} \right]}^{m}}={}^{m}{{C}_{0}}-{}^{m}{{C}_{1}}{{(1+x)}^{n}}+{}^{m}{{C}_{2}}{{(1+x)}^{2n}}+\,........$
Substituting the above equation in equation (1),
$LHS=$ Coefficient of ${{x}^{m}}$ in $\left\{ 1-{{\left[ 1-{{(1+x)}^{n}} \right]}^{m}} \right\}$
Now expanding using ${{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}\,.........\,$
$LHS=$ Coefficient of ${{x}^{m}}$ in $\left\{ 1-{{\left[ 1-1-{}^{n}{{C}_{1}}x-{}^{n}{{C}_{1}}{{x}^{2}}-\,...\,.... \right]}^{m}} \right\}$
Now taking -1 common,
$LHS=$ Coefficient of ${{x}^{m}}$ in $\left\{ 1+{{(-1)}^{m-1}}{{\left[ {}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+\,......... \right]}^{m}} \right\}$
Now taking ${{x}^{m}}$ common from the inside of the expansion,
$LHS=$ Coefficient of ${{x}^{m}}$ in $\left\{ 1+{{(-1)}^{m-1}}{{x}^{m}}{{\left[ {}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}x+\,...\,....... \right]}^{m}} \right\}$
=\[{{(-1)}^{m-1}}{{\left( {}^{n}{{C}_{1}} \right)}^{m}}\]
We know that the value of \[{}^{n}{{C}_{1}}\] is n,
$LHS=$ \[{{(-1)}^{m-1}}{{n}^{m}}\]
$=RHS$
${}^{m}{{C}_{1}}{}^{n}{{C}_{m}}-{}^{m}{{C}_{2}}{}^{2n}{{C}_{m}}+{}^{m}{{C}_{3}}{}^{3n}{{C}_{m}}..............\,\,=\,\,{{(-1)}^{m-1}}{{n}^{m}}$
Hence proved.

Note:

The formula ${{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{0}}x+{}^{n}{{C}_{0}}{{x}^{2}}+\,......\,+{}^{n}{{C}_{n}}{{x}^{n}}$ is a powerful tool, which is very flexible and with simple alteration, is used to obtain several other equation.
Nested binomial expansions have to be solved separately. By nested binomial expansions, we mean expansions inside expansions. In this question, ${{\left[ 1-{{(1+x)}^{n}} \right]}^{m}}$ is such an expansion.