Question

State and prove the multinomial theorem.

Answer 1

We have to state and prove the multinomial theorem for this question. For that, we will first state the theorem which is given as follows:
“For a positive integer k and non-negative integer n,
${{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{k}} \right)}^{n}}=\sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{k}}=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{k}}}{\overset{n}{\mathop{{}}}}\, \right)}\prod\limits_{j=1}^{k}{{{x}_{j}}^{{{b}_{j}}}}$
Where, $\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{k}}}{\overset{n}{\mathop{{}}}}\, \right)$ is given as:
$\left(\underset{{{b}_{1}},{{b}_{2}},...,{{b}_{k}}}{\overset{n}{\mathop{{}}}}\, \right)=\dfrac{n!}{{{b}_{1}}!{{b}_{2}}!...{{b}_{k}}!}$ ”
Then we will prove this by the principle of mathematical induction. For this, we will first prove that the multinomial theorem is true for k=1. Then we will assume that this theorem is true for k=m and then we will try to prove that the theorem is true for k=m+1 by using the equation we got from assuming that the theorem is true for k=m. In this, we will take the last two terms of the series as one single term and use the k=m equation as there will be then m terms. Then we will solve it and try to get it in the form of RHS when we will keep k=m+1. Hence, our theorem will be proved.

Complete step-by-step solution:
We need to state and prove the multinomial theorem.
For this, we will first give its statement.
Multinomial theorem is stated as:
“For a positive integer k and non-negative integer n,
${{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{k}} \right)}^{n}}=\sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{k}}=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{k}}}{\overset{n}{\mathop{{}}}}\, \right)}\prod\limits_{j=1}^{k}{{{x}_{j}}^{{{b}_{j}}}}$
Where, $\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{k}}}{\overset{n}{\mathop{{}}}}\, \right)$ is given as:
$\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{k}}}{\overset{n}{\mathop{{}}}}\, \right)=\dfrac{n!}{{{b}_{1}}!{{b}_{2}}!...{{b}_{k}}!}$ ”
Now, we will try to prove this by the method of mathematical induction in k.
Now, let us see if the multinomial theorem is true for k=1 or not.
Taking k=1, we get the LHS as:
$\begin{aligned} & LHS={{\left( {{x}_{1}}+{{x}_{2}}+...+{{x}_{k}} \right)}^{n}} \\\ & \Rightarrow LHS={{\left( {{x}_{1}} \right)}^{n}} \\\ & \Rightarrow LHS={{x}_{1}}^{n} \\\ \end{aligned}$
Now, when k=1, the RHS will be:
$\begin{aligned} & \sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{k}}=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{k}}}{\overset{n}{\mathop{{}}}}\, \right)}\prod\limits_{j=1}^{k}{{{x}_{j}}^{{{b}_{j}}}} \\\ & \Rightarrow \sum\limits_{{{b}_{1}}=n}{\left( \underset{{{b}_{1}}}{\overset{n}{\mathop{{}}}}\, \right)}\prod\limits_{j=1}^{1}{{{x}_{j}}^{{{b}_{j}}}} \\\ \end{aligned}$
From this, we can see that ${{b}_{1}}=n$
Thus, we get:
$\begin{aligned} & RHS=\sum\limits_{{{b}_{1}}=n}{\left( \underset{{{b}_{1}}}{\overset{n}{\mathop{{}}}}\, \right)}\prod\limits_{j=1}^{I}{{{x}_{j}}^{{{b}_{j}}}} \\\ & \Rightarrow RHS=\sum\limits_{n}{\left( \underset{n}{\overset{n}{\mathop{{}}}}\, \right)}{{\left( {{x}_{1}} \right)}^{{{b}_{1}}}} \\\ & \Rightarrow RHS=\dfrac{n!}{n!}{{\left( {{x}_{1}} \right)}^{n}} \\\ & \Rightarrow RHS={{x}_{1}}^{n} \\\ \end{aligned}$
Thus, we can see that LHS=RHS.
Hence, the multinomial theorem is true for k=1.
Now, let us assume that the multinomial theorem is true for k=m where m is any positive integer.
Thus, we can say that:
${{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{m}} \right)}^{n}}=\sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m}}=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{m}}}{\overset{n}{\mathop{{}}}}\, \right)}\prod\limits_{j=1}^{m}{{{x}_{j}}^{{{b}_{j}}}}$
Now, that we have assumed that the multinomial theorem is true for k=m, we will see that if it is true for k=m+1 or not.
Putting k=m+1 in multinomial theorem we get the LHS as:
${{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{m}}+{{x}_{m+1}} \right)}^{n}}$
Now, let us assume that ${{x}_{m}}+{{x}_{m+1}}$ is a single term.
Thus, ${{b}_{m}}+{{b}_{m+1}}$ will also be a single term.
Let us assume this term to be ‘M’.
As a result, the number of terms we will get will be:
m+1-1=m
Thus, we can write the multinomial theorem as:
$\Rightarrow {{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{m-1}}+({{x}_{m}}+{{x}_{m+1}}) \right)}^{n}}=\sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m-1}}+M=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{m-1}},M}{\overset{n}{\mathop{{}}}}\, \right)}\prod\limits_{j=1}^{m-1}{{{x}_{j}}^{{{b}_{j}}}.{{\left( {{x}_{m}}+{{x}_{m+1}} \right)}^{M}}}$
Now, we can write the expansion of ${{\left( {{x}_{m}}+{{x}_{m+1}} \right)}^{M}}$ by using the binomial theorem.
We know that the expansion of ${{\left( a+b \right)}^{n}}$ is written as:
$\sum\limits_{r+n-r=n}{\left( \underset{r,n-r}{\overset{n}{\mathop {}}}\, \right){{a}^{n-r}}{{b}^{r}}}$
Thus, ${{\left( {{x}_{m}}+{{x}_{m+1}} \right)}^{M}}$ is given as:
$\begin{aligned} & {{\left( {{x}_{m}}+{{x}_{m+1}} \right)}^{M}} \\\ & \sum\limits_{{{b}_{m}}+M-{{b}_{m}}=M}{\left( \underset{{{b}_{m}},M-{{b}_{m}}}{\overset{M}{\mathop {}}}\, \right){{x}_{m}}^{{{b}_{m}}}{{x}_{m+1}}^{M-{{b}_{m}}}} \\\ \end{aligned}$
Now, we already established above that ${{b}_{m}}+{{b}_{m+1}}=M$
Thus, we can say that:
${{b}_{m+1}}=M-{{b}_{m}}$
Thus, we can write ${{\left( {{x}_{m}}+{{x}_{m+1}} \right)}^{M}}$ as:
$\sum\limits_{{{b}_{m}}+{{b}_{m+1}}=M}{\left( \underset{{{b}_{m}},{{b}_{m+1}}}{\overset{M}{\mathop {}}}\, \right){{x}_{m}}^{{{b}_{m}}}{{x}_{m+1}}^{{{b}_{m+1}}}}$
Thus, putting this value in the expansion, we get:

& {{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{m-1}}+({{x}_{m}}+{{x}_{m+1}}) \right)}^{n}}=\sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m-1}}+M=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{m-1}},M}{\overset{n}{\mathop{{}}}}\, \right)}\prod\limits_{j=1}^{m-1}{{{x}_{j}}^{{{b}_{j}}}.{{\left( {{x}_{m}}+{{x}_{m+1}} \right)}^{M}}} \\\ & \Rightarrow {{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{m-1}}+({{x}_{m}}+{{x}_{m+1}}) \right)}^{n}}=\sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m-1}}+M=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{m-1}},M}{\overset{n}{\mathop{{}}}}\, \right)}\prod\limits_{j=1}^{m-1}{{{x}_{j}}^{{{b}_{j}}}.\sum\limits_{{{b}_{m}}+{{b}_{m+1}}=M}{\left( \underset{{{b}_{m}},{{b}_{m+1}}}{\overset{M}{\mathop {}}}\, \right){{x}_{m}}^{{{b}_{m}}}{{x}_{m+1}}^{{{b}_{m+1}}}}} \\\ & \Rightarrow {{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{m-1}}+({{x}_{m}}+{{x}_{m+1}}) \right)}^{n}}=\sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m-1}}+M=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{m-1}},M}{\overset{n}{\mathop{{}}}}\, \right)}.\sum\limits_{{{b}_{m}}+{{b}_{m+1}}=M}{\left( \underset{{{b}_{m}},{{b}_{m+1}}}{\overset{M}{\mathop {}}}\, \right)}.\prod\limits_{j=1}^{m-1}{{{x}_{j}}^{{{b}_{j}}}}.{{x}_{m}}^{{{b}_{m}}}{{x}_{m+1}}^{{{b}_{m+1}}} \\\ \end{aligned}$$ Now, we will find the value of $$\sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m-1}}+M=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{m-1}},M}{\overset{n}{\mathop{{}}}}\, \right)}.\sum\limits_{{{b}_{m}}+{{b}_{m+1}}=M}{\left( \underset{{{b}_{m}},{{b}_{m+1}}}{\overset{M}{\mathop {}}}\, \right)}$$ by expanding both these sigmas. Expanding both the sigmas, we get: $$\begin{aligned} & \sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m-1}}+M=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{m-1}},M}{\overset{n}{\mathop{{}}}}\, \right)}.\sum\limits_{{{b}_{m}}+{{b}_{m+1}}=M}{\left( \underset{{{b}_{m}},{{b}_{m+1}}}{\overset{M}{\mathop {}}}\, \right)} \\\ & \Rightarrow \sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m-1}}+M=n}{\dfrac{n!}{{{b}_{1}}!{{b}_{2}}!...{{b}_{m-1}}!M!}}.\sum\limits_{{{b}_{m}}+{{b}_{m+1}}=M}{\dfrac{M!}{{{b}_{m}}!{{b}_{m+1}}!}} \\\ & \Rightarrow \sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m}}+{{b}_{m+1}}=n}{\dfrac{n!}{{{b}_{1}}!{{b}_{2}}!...{{b}_{m-1}}!M!}.\dfrac{M!}{{{b}_{m}}!{{b}_{m+1}}}} \\\ & \Rightarrow \sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m}}+{{b}_{m+1}}=n}{\dfrac{n!}{{{b}_{1}}!{{b}_{2}}!...{{b}_{m}}!{{b}_{m+1}}!}} \\\ & \Rightarrow \sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m}}+{{b}_{m+1}}=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{m}},{{b}_{m+1}}}{\overset{n}{\mathop {}}}\, \right)} \\\ \end{aligned}$$ Now, we will solve $$\prod\limits_{j=1}^{m-1}{{{x}_{j}}^{{{b}_{j}}}}.{{x}_{m}}^{{{b}_{m}}}{{x}_{m+1}}^{{{b}_{m+1}}}$$. Now, we can write $$\prod\limits_{j=1}^{m-1}{{{x}_{j}}^{{{b}_{j}}}}.{{x}_{m}}^{{{b}_{m}}}{{x}_{m+1}}^{{{b}_{m+1}}}$$ as: $$\begin{aligned} & \prod\limits_{j=1}^{m-1}{{{x}_{j}}^{{{b}_{j}}}}.{{x}_{m}}^{{{b}_{m}}}{{x}_{m+1}}^{{{b}_{m+1}}} \\\ & \Rightarrow {{x}_{1}}^{{{b}_{1}}}.{{x}_{2}}^{{{b}_{2}}}.{{x}_{3}}^{{{b}_{3}}}...{{x}_{m-1}}^{{{b}_{m-1}}}.{{x}_{m}}^{{{b}_{m}}}.{{x}_{m+1}}^{{{b}_{m+1}}} \\\ & \Rightarrow \prod\limits_{j=1}^{m+1}{{{x}_{j}}^{{{b}_{j}}}} \\\ \end{aligned}$$ Now, putting these values in the expansion we get: $$\begin{aligned} & {{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{m-1}}+({{x}_{m}}+{{x}_{m+1}}) \right)}^{n}}=\sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m-1}}+M=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{m-1}},M}{\overset{n}{\mathop{{}}}}\, \right)}.\sum\limits_{{{b}_{m}}+{{b}_{m+1}}=M}{\left( \underset{{{b}_{m}},{{b}_{m+1}}}{\overset{M}{\mathop {}}}\, \right)}.\prod\limits_{j=1}^{m-1}{{{x}_{j}}^{{{b}_{j}}}}.{{x}_{m}}^{{{b}_{m}}}{{x}_{m+1}}^{{{b}_{m+1}}} \\\ & {{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{m}}+{{x}_{m+1}} \right)}^{n}}=\sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m}}+{{b}_{m+1}}=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{m}},{{b}_{m+1}}}{\overset{n}{\mathop {}}}\, \right)}\prod\limits_{j=1}^{m+1}{{{x}_{j}}^{{{b}_{j}}}} \\\ \end{aligned}$$ Now, if put k=m+1 in the RHS of the multinomial theorem, we get: $\begin{aligned} & RHS=\sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{k}}=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{k}}}{\overset{n}{\mathop{{}}}}\, \right)}\prod\limits_{j=1}^{k}{{{x}_{j}}^{{{b}_{j}}}} \\\ & \Rightarrow RHS=\sum\limits_{{{b}_{1}}+{{b}_{2}}+...+{{b}_{m}}+{{b}_{m+1}}=n}{\left( \underset{{{b}_{1}},{{b}_{2}},...,{{b}_{m}},{{b}_{m+1}}}{\overset{n}{\mathop{{}}}}\, \right)}\prod\limits_{j=1}^{m+1}{{{x}_{j}}^{{{b}_{j}}}} \\\ \end{aligned}$ Hence, we can see that LHS=RHS. Thus, we can see that the multinomial theorem is true for k=m+1 if it is true for k=m and since k=1 is true, thus, by the principle of mathematical induction, we can say that the multinomial theorem is true for all values k such that k is a natural number. Hence, the multinomial theorem is proved. **Note:** Solve the sigmas and the products very carefully as they can prove to be really confusing and hence our theorem wouldn’t be proved. It is important to solve the sigmas separately and the products separately as that is the way in which making mistakes is the least likely. Also, since it is proved by the principle of mathematical induction, remember to mention all the statements as they are an important part of the process.