Question

Prove the binomial theorem using mathematical induction.

Short answer

By the principle of mathematical induction, \(P(n)\) is true for all positive integers \(n\). The required expression is \({(a + b)^n} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right)} {a^{n - k}}{b^k}\).

Step-by-step solution

Given variables and integer

\(a\)and \(b\) are variables

\({\bf{n}}\) is a positive integer

Definition of Binomial theorem

Binomial theorem, states that for any positive integer\(n\), the\(n\)th power of the sum of two numbers\(a\)and\(b\)may be expressed as the sum of\(n + 1\)terms of the form.

Proof by induction

Let \(P(n)\) be the statement.

\({(a + b)^n} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right)} {a^{n - k}}{b^k}\)

Basis step:

\(n = 1\)

\(\begin{array}{l}{(a + b)^n} = {(a + b)^1}\\{(a + b)^n} = a + b\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right)} {a^{n - k}}{b^k}\\{(a + b)^n} = \sum\limits_{k = 0}^1 {\left( {\begin{array}{*{20}{l}}1\\k\end{array}} \right)} {a^{1 - k}}{b^k}\end{array}\)

\(\begin{array}{l}{(a + b)^n} = {a^1}{b^0} + {a^0}{b^1}\\{(a + b)^n} = a + b\end{array}\)

Thus \(P(1)\) is true.

Inductive step:

Let \(P(m)\) be true.

\({(a + b)^m} = \sum\limits_{k = 0}^m {\left( {\begin{array}{*{20}{c}}m\\k\end{array}} \right)} {a^{m - k}}{b^k}\)

Use distributive property and Pascal’s identity

prove that \(P(m + 1)\) is true:

\({(a + b)^{m + 1}} = (a + b){(a + b)^m}\)

Since \(P(k)\) is true.

\({(a + b)^{m + 1}} = (a + b)\left( {\sum\limits_{k = 0}^m {\left( {\begin{array}{*{20}{c}}m\\k\end{array}} \right)} {a^{m - k}}{b^k}} \right)\)

Use distributive property:

\(\begin{array}{l}{(a + b)^{m + 1}} = \sum\limits_{k = 0}^m {\left( {\begin{array}{*{20}{c}}m\\k\end{array}} \right)} {a^{m - k + 1}}{b^k} + \sum\limits_{k = 0}^m {\left( {\begin{array}{*{20}{c}}m\\k\end{array}} \right)} {a^{m - k}}{b^{k + 1}}\\{(a + b)^{m + 1}} = \sum\limits_{k = 0}^m {\left( {\begin{array}{*{20}{c}}m\\k\end{array}} \right)} {a^{m - k + 1}}{b^k} + \sum\limits_{k = 1}^{m + 1} {\left( {\begin{array}{*{20}{c}}m\\{k - 1}\end{array}} \right)} {a^{m - (k - 1)}}{b^k}\end{array}\)

\(\begin{array}{l}{(a + b)^{m + 1}} = \left( {\begin{array}{*{20}{c}}m\\0\end{array}} \right){a^{m + 1}}{b^0} + \sum\limits_{k = 1}^m {\left( {\left( {\begin{array}{*{20}{c}}m\\k\end{array}} \right) + \left( {\begin{array}{*{20}{c}}m\\{k - 1}\end{array}} \right)} \right)} {a^{m - k + 1}}{b^k} + \left( {\begin{array}{*{20}{c}}m\\m\end{array}} \right){a^0}{b^{m + 1}}\\{(a + b)^{m + 1}} = {a^{m + 1}} + \sum\limits_{k = 1}^m {\left( {\left( {\begin{array}{*{20}{c}}m\\k\end{array}} \right) + \left( {\begin{array}{*{20}{c}}m\\{k - 1}\end{array}} \right)} \right)} {a^{m - k + 1}}{b^k} + {b^{m + 1}}\end{array}\)

Use Pascal's identity:

\(\begin{array}{l}\left( {\begin{array}{*{20}{c}}{n + 1}\\k\end{array}} \right) = \left( {\begin{array}{*{20}{c}}n\\{k - 1}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{n + 1}\\k\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{m + 1}\\0\end{array}} \right){a^{m + 1}}{b^0} + \sum\limits_{k = 1}^m {\left( {\begin{array}{*{20}{c}}{m + 1}\\k\end{array}} \right)} {a^{m - k + 1}}{b^k} + \left( {\begin{array}{*{20}{c}}{m + 1}\\{m + 1}\end{array}} \right){a^0}{b^{m + 1}}\\\left( {\begin{array}{*{20}{c}}{n + 1}\\k\end{array}} \right) = \sum\limits_{k = 0}^{m + 1} {\left( {\begin{array}{*{20}{c}}{m + 1}\\k\end{array}} \right)} {a^{m - k + 1}}{b^k}\end{array}\)

Thus \(P(k + 1)\) is true.