Question

In the binomial expansion of ${\left( {1 + x} \right)^{m + n}}$ , prove that the coefficient of ${x^m}$ and ${x^n}$ are equal.

Answer 1

Hint: In order to solve the problem first use the formula for the general coefficient of a binomial expansion to find the coefficient of both the given terms in the problem. Further prove that both of them are equal by using the formula for combination terms.

Complete step by step answer:
Given that the binomial expansion term is ${\left( {1 + x} \right)^{m + n}}$
We have to prove that the coefficients of ${x^m}$ and ${x^n}$ are the same.
So, first let us find the coefficients of ${x^m}$ and ${x^n}$ in the expansion.
As we know that for any general binomial expansion term ${\left( {1 + x} \right)^k}$ the general coefficient of ${x^r}$ is given by \[{}^k{C_r}\] .
Using the above formula let us find the coefficients of ${x^m}$ and ${x^n}$ .
The coefficient of ${x^m}$ is \[{}^{m + n}{C_m}\] .
Similarly the coefficient of ${x^n}$ is \[{}^{m + n}{C_n}\] .
Now we have the coefficients of both the terms we have to prove they are equal.
As we know the general formula for combination term is:
\[{}^k{C_r} = {}^k{C_{k - r}}\]
Using the above formula let us manipulate the coefficient of ${x^n}$ . So, we have:
\[
   \Rightarrow {}^{m + n}{C_n} = {}^{m + n}{C_{\left( {m + n} \right) - n}} \\
   \Rightarrow {}^{m + n}{C_n} = {}^{m + n}{C_m} \\
 \]
This is the same as the coefficient of ${x^m}$ .
Hence, the coefficient of ${x^m}$ and ${x^n}$ are equal.

Note: In order to solve such problems, students must remember the formulas connecting different combination terms. Students must also remember the general formula for the expansion of a binomial series and also the general terms of the binomial expansion. The binomial coefficient used above in the given problem is the number of ways out of possibilities to choose unordered results, also known as combination or combinatorial sum. The symbols are used to denote a coefficient of binomials and are often read as "choose."