Binomial Theorem and Binomial Coefficient

The binomial theorem or binomial expansion desribes the algebraic expansion of powers of a binomial. The teorem states that it is possible to expand the polynomial \((x+y)^n\) into a sum that includes terms \(ax^by^c\). The coefficient \(a\) in term \(ax^by^c\) is binomial coefficient while \(b\) and \(c\) are nonegative integers whose sum gives number \(n\) i.e. \(b+c = n\). For positive integer n: \begin{eqnarray}(x+y)^n &=& \binom{n}{0}x^ny^0 + \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2} y^2+...+\binom{n}{n-1}x^1 y^{n-1} + \binom{n}{n} x^0 y^n\\ (x+y)^n &=& \sum_{k = 0}^n \binom{n}{k} x^{n-k}b^k. \end{eqnarray} The properties of binomial coefficient are:

• the developt binomial has \(n+1\) members,
• exponents with base \(x\) decreasse from \(n\) to 0 while the exponennt with base \(y\) increase from 0 to \(n\), their sum in every member is equal to n.
• if we observe the formula from left and right side at the same time we can see that binomial coefficient are symmetric (the value is equal).

The binomial coefficient is natural number that is calculated using the formula which can be written as $$ \binom{n}{k} = \frac{n!}{(n-k)!k!}, \quad n \in \mathbf{N}, k \in N \cup{0}, k\leq n.$$ The special case of binomial coefficient when \(k = 0\) then binomial coefficient \(\binom{n}{0}\) by definition is equal to 1. For smaller numbers \(n\) and \(k\) the binomial coefficient can be calculated without formula, so that in the denominator we write the product of all numbers from 1 to \(k\), and in the numerator we write the product of as many numbers as there are in the denominator, starting with \(n\), in descending order. For example calcualte the value of binomial coefficient \(\binom{10}{6}\). $$ \require{cancel} \binom{10}{6} = \frac{\cancelto{5}{10}\cdot \cancelto{3}{9} \cdot \cancelto{2}{8} \cdot 7 \cdot \cancel{6} \cdot \cancel{5}}{1\cdot \cancel{2} \cdot \cancel{3} \cdot \cancel{4} \cdot \cancel{5} \cdot \cancel{6}} = 5\cdot 3 \cdot 2 \cdot 7 = 210$$ Properties of binomial coefficeints Two basic properties are:

• the symetry property $$ \binom{n}{k} = \binom{n}{n-k} $$ For example \(\binom{14}{10} = \binom{14}{4}\)
• the Pascal's triangle property $$\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k} $$ For example: \(\binom{7}{3} + \binom{7}{4} = \binom{8}{4}\)

Binomial theorem and Binomial coefficients - Examples

• Example 1 Write the following expression \((x+y)^2\) in the expanded form using binomial theorem. $$\begin{eqnarray}(x+y)^2 &=& x^2 + 2x^{2-1} y + y^2 = x^2 + 2xy + y^2\end{eqnarray}$$
Example 2 Write the following expression \((x+y)^3\) in the expanded form using binomial theorem. $$ \begin{eqnarray}(x+y)^3 &=& x^3 + 3x^{3-1}y + \frac{3(3-1)}{2!}x^{3-2}y^2 + \frac{3(3-1)(3-2)}{3!}x^{3-3}y^3 \\\nonumber (x+y)^3 &=& x^3 + 3x^2 y + 3xy^2 + y^3\end{eqnarray}$$ Example 3 Write the following expression \((x+y)^4\) in the expanded form using binomial theorem. $$ \begin{eqnarray}(x+y)^4 &=& x^4 + 4x^{4-1}y + \frac{4(4-1)}{2!}x^{4-2}y^2 + \frac{4(4-1)(4-2)}{3!}x^{4-3}y^3 + \frac{4(4-1)(4-2)(4-3)}{4!}x^{4-4}y^4 \\ \nonumber (x+y)^4 &=& x^4 + 4x^3y+ 6x^2y^2 + 8xy^3 + y^4\end{eqnarray}$$ Example 4 Write the following expression \((x+y)^6\) in expanded form. $$\begin{eqnarray}(x+y)^6 &=& x^6 + 6x^5y + \frac{6(6-1)}{2!}x^{6-2}y^2 + \frac{6(6-1)(6-2)}{3!}x^{6-3}y^3 \\ \nonumber &+& \frac{6(6-1)(6-2)(6-3)}{4!}x^{6-4}y^4 + \frac{6(6-1)(6-2)(6-3)(6-4)}{5!}x^{6-5}y^5 \\ \nonumber &+&\frac{6(6-1)(6-2)(6-2)(6-3)(6-4)(6-5)}{6!}y^6\\\nonumber (x+y)^6 &=& x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3+15x^2 y^4 + 6xy^5 + y^6\end{eqnarray}$$
• Example 5 Solve the equation \(\binom{n}{6} = \binom{n}{5}\)
  Solution $$\begin{eqnarray}\frac{n!}{(n-6)!6!} &=&\frac{n!}{(n-5)!5!}\\ \nonumber \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)!}{(n-6)!6!} &=& \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)!}{(n-5)!5!}\\\nonumber \frac{n-5}{6\cdot 5!} &=& \frac{1}{5!} \\\nonumber n-5 &=& 6 \\\nonumber n &=& 11 \end{eqnarray}$$
• Example 6 Solve the equation\(\binom{n}{4} + \binom{n}{5} = \binom{17}{5}\)
  Solution $$ \begin{eqnarray} \binom{n}{4} + \binom{n}{5} &=& \binom{17}{5} \\ \nonumber \end{eqnarray}$$ In this example the \(n\) can be found by applying Pascal's trinagle property. The formula of Pascal's triangle can be written as $$\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k} $$. By comparing Pascals triangle property with our example we can conclude that \(k = 5\) and \(n+1 = 17 \Rightarrow n = 16\)
• Example 7 Which term in order of development of binomial \(\left(x^2 + y^3\right)^5\) contains \(x^6y^6\)?
  Solution The general term of development $$ A_{k+1} = \binom{5}{k}(x^2)^{5-k}(y^3)^k = \binom{5}{k}x^{10-2k}y^{3k}$$ To determine \(k\) we equate the exponents with the bases \begin{eqnarray} 10-2k &=& 6 \Rightarrow -2k = -4/:(-2) \Rightarrow k = 2\\ 3k &=& 6/:3 \Rightarrow k = 2\end{eqnarray} The \(x^6y^6\) is the third member.
• Example 8 Determine the coefficient in development of binomial \(\left(3+4x\right)^6\) which contains \(x^3\)
  Solution\begin{eqnarray}A_{k+1} &=& \binom{6}{k}3^{6-k}(4x)^k = \binom{6}{k}3^{6-k}4^kx^k\\ x^k &=& x^3 \Rightarrow k = 3\\ \binom{6}{3}&\cdot & 3^3\cdot 4^3 \cdot x^3 = 20\cdot 27 \cdot 64\cdot x^3 = 34560 x^3\end{eqnarray} The coefficient which is next to \(x^3\) is 34560.
• Example 9 If a binomial coefficient of third member of binomial \(\left(9x - \frac{1}{\sqrt{3x}}\right)^n\)is equal to 105, determine the thirtheenth member.
  Solution From description of the problem we know that the third member is equal to 105. \begin{eqnarray}\require{cancel} \binom{n}{2} &=& 105 \\ \frac{n!}{(n-2)! 2!} &=& 105 \\ \frac{n(n-1)\cancel{(n-2)!}}{\cancel{(n-2)!}2} &=& 105/\cdot 2\\ n(n-1) &=& 210 \\ n^2 - n - 210 &=& 0 \\ n_{1,2} &=& \frac{1 \pm \sqrt{1 + 4*210}}{2} \\ n_{1,2} &=& \frac{1 \pm 29}{2} \\ n_1 &=& 15 \\ \cancel{n_2 = -14} \end{eqnarray} The thirtheen member is calculated using formula for general member \begin{eqnarray}A_{13} &=& \binom{15}{12} (9x)^3 \left((3x)^{-\frac{1}{2}}\right)^{12}\\ A_{13} &=& 455 \cdot 3^6x^3 (3x)^{-6}\\ A_{13} &=& 455 x^{-3}\end{eqnarray}
• Example 10 Calculate using binomial formula \(0.95^5\) of four decimal places.
  Solution \begin{eqnarray}0.95^5 &=& (1-0.05)^5\\ 0.95^5 &=& 1*5\cdot 0.05 + 10\cdot 0.05^2 - 10\cdot 0.05^3 + 5\cdot 0.05^4 - 0.05^5 = 0.7738\end{eqnarray}