Question

State and prove a multinomial theorem?

Answer 1

Hint: Multinomial theorem is the generalization of binomial theorem. Binomial theorem is given as \[{\left( {a + b} \right)^n} = {}^n{C_0}{\left( a \right)^n}{\left( b \right)^0} + {}^n{C_1}{\left( a \right)^{n - 1}}{\left( b \right)^1} + {}^n{C_2}{\left( a \right)^{n - 2}}{\left( b \right)^2} + ..... + {}^n{C_n}{\left( a \right)^0}{\left( b \right)^n}\].

Complete step by step solution:
Step 1: We have to state the multinomial theorem. It is the generalization of the binomial theorem. It describes how to expand a power of a sum in terms of powers of the terms in that sum. It states that “For any positive integer $m$ and any non – negative integer $n$ the sum of $m$ terms raised to power $n$ is expanded as
${\left( {{x_1} + {x_2} + ..... + {x_m}} \right)^n} = \sum\limits_{{k_1} + {k_2} + .... + {k_m} = n} {\left( {\begin{array}{*{20}{c}}
  n \\
  {{k_1},{k_2},....,{k_m}}
\end{array}} \right)} \prod\limits_{t = 1}^m {x_t^{{k_t}}} $
Where $\left( {\begin{array}{*{20}{c}}
  n \\
  {{k_1},{k_2},....,{k_m}}
\end{array}} \right) = \dfrac{{n!}}{{{k_1}!{k_2}!....{k_m}!}}$ is a multinomial coefficient. “
Here the important thing to keep in mind is that the sum is taken over all combinations of non negative integer indices ${k_1}$ to ${k_m}$ is such that the sum of all ${k_i}$ is $n$ .That is, for each term in the expansion, the exponent of the ${x_i}$ must adds up to $n$ .
When the value of $n = 2$ the multinomial theorem is converted into binomial theorem.
 Step 2: Now, we can prove multinomial theorem by using binomial theorem and rule of mathematical induction. The expression holds for $m = 1$ as LHS and RHS are equal.
Now, let us consider that above expression is true for the value of $m$ also.
Now we will prove the expression for $m + 1$ . For that, writing the above expression for $m + 1$ terms, we get
${\left( {{x_1} + {x_2} + ..... + {x_m} + {x_{m + 1}}} \right)^n}$
Separating the last two terms, we get
\[
   \Rightarrow {\left( {{x_1} + {x_2} + ..... + {x_{m - 1}} + \left( {{x_m} + {x_{m + 1}}} \right)} \right)^n} \\
   \Rightarrow \sum\limits_{{k_1} + {k_2} + .... + {k_{m - 1}} + K = n} {\left( {\begin{array}{*{20}{c}}
  n \\
  {{k_1},{k_2},....,{k_{m - 1}},K}
\end{array}} \right)} \prod\limits_{t = 1}^{m - 1} {x_t^{{k_t}}} \times {\left( {{x_m} + {x_{m + 1}}} \right)^K} \\
 \]
Now applying the binomial theorem for the last factor, we get
\[ \Rightarrow \sum\limits_{{k_1} + {k_2} + .... + {k_{m - 1}} + K = n} {\left( {\begin{array}{*{20}{c}}
  n \\
  {{k_1},{k_2},....,{k_{m - 1}},K}
\end{array}} \right)} \left( {x_1^{{k_1}}x_2^{{k_2}}....x_{m - 1}^{{k_{m - 1}}}} \right) \times \sum\limits_{{k_m} + {k_{m + 1}} = K} {\left( {\begin{array}{*{20}{c}}
  K \\
  {{k_m},{k_{m + 1}}}
\end{array}} \right)} \left( {x_m^{{k_m}}x_{m + 1}^{{k_{m + 1}}}} \right)\]
Step 3: Now in the above step, the value of
\[
  \left( {\begin{array}{*{20}{c}}
  n \\
  {{k_1},{k_2},....,{k_{m - 1}},K}
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  K \\
  {{k_m},{k_{m + 1}}}
\end{array}} \right) \\
   = \dfrac{{n!}}{{{k_1}!{k_2}!....{k_{m - 1}}!K!}} \times \dfrac{{K!}}{{{k_m}!{k_{m + 1}}!}} \\
   = \dfrac{{n!}}{{{k_1}!{k_2}!....{k_{m - 1}}!{k_m}!{k_{m + 1}}!}} \\
   = \left( {\begin{array}{*{20}{c}}
  n \\
  {{k_1},{k_2},....,{k_{m - 1}},{k_m},{k_{m + 1}}}
\end{array}} \right) \\
 \]
Substituting the value, we get
\[ \Rightarrow \sum\limits_{{k_1} + {k_2} + .... + {k_{m - 1}} + {k_m} + {k_{m + 1}} = n} {\left( {\begin{array}{*{20}{c}}
  n \\
  {{k_1},{k_2},....,{k_{m - 1}},{k_m},{k_{m + 1}}}
\end{array}} \right)} \left( {x_1^{{k_1}}x_2^{{k_2}}....x_{m - 1}^{{k_{m - 1}}}x_m^{{k_m}}x_{m + 1}^{{k_{m + 1}}}} \right)\]
Hence proved.

Note: The induction hypothesis is proved for $n = 1$ . Then we assume that the statement is true for $n$ terms. Then by using this assumed result, we proved the statement for $n + 1$ terms.