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Question: How do I find the constant term of a binomial expansion?...

How do I find the constant term of a binomial expansion?

Explanation

Solution

In the above question, you were asked to find the constant term of binomial expression. According to the formula of the binomial theorem that is (x+y)n{{(x+y)}^{n}} , the term yn{{y}^{n}} is always constant. You will see how this is a constant term. So, let us see how we can solve this problem.

Complete step by step solution:
We will see what do we get from the expansion of the binomial formula which is (x+y)n{{(x+y)}^{n}}
⇒(x+y)n=(n 0 ).xn+(n 1 ).xn−1.y1....+(n k ).xn−k.yk+....+(n n ).yn=∑k=0n.(n k ).xn−k.yk\Rightarrow {{(x+y)}^{n}}=(\begin{matrix} n \\\ 0 \\\ \end{matrix}).{{x}^{n}}+(\begin{matrix} n \\\ 1 \\\ \end{matrix}).{{x}^{n-1}}.{{y}^{1}}....+(\begin{matrix} n \\\ k \\\ \end{matrix}).{{x}^{n-k}}.{{y}^{k}}+....+(\begin{matrix} n \\\ n \\\ \end{matrix}).{{y}^{n}}=\sum\limits_{k=0}^{n}{.(\begin{matrix} n \\\ k \\\ \end{matrix}).{{x}^{n-k}}.{{y}^{k}}}
where x,y∈R,x,y\in \mathbb{R}, k,n∈Nk,n\in \mathbb{N} and (n k )(\begin{matrix} n \\\ k \\\ \end{matrix}) denotes combinations of n things taken k at a time. So we have 2 cases
1st case: When the terms of the binomial are a constant and a variable like
⇒(x+c)n=(n 0 )⋅xn+(n 1 )⋅xn−1⋅c1+...+(n k )⋅xn−k⋅ck+...+(n n )⋅cn\Rightarrow {{(x+c)}^{n}}=(\begin{matrix} n \\\ 0 \\\ \end{matrix})\cdot {{x}^{n}}+(\begin{matrix} n \\\ 1 \\\ \end{matrix})\cdot {{x}^{n-1}}\cdot {{c}^{1}}+...+(\begin{matrix} n \\\ k \\\ \end{matrix})\cdot {{x}^{n-k}}\cdot {{c}^{k}}+...+(\begin{matrix} n \\\ n \\\ \end{matrix})\cdot {{c}^{n}}
Here the constant term is (n n )⋅cn(\begin{matrix} n \\\ n \\\ \end{matrix})\cdot {{c}^{n}} and its product is also constant.
2nd Case: When the terms of the binomial are a variable and the ratio of that variable like
⇒(n k )⋅xn−k.(cx)k=(n k )⋅xn−k⋅ck.1xk=((n n )⋅ck).xn−kxk=((n k )⋅ck).xn−2k\Rightarrow (\begin{matrix} n \\\ k \\\ \end{matrix})\cdot {{x}^{n-k}}.{{(\dfrac{c}{x})}^{k}}=(\begin{matrix} n \\\ k \\\ \end{matrix})\cdot {{x}^{n-k}}\cdot {{c}^{k}}.\dfrac{1}{{{x}^{k}}}=((\begin{matrix} n \\\ n \\\ \end{matrix})\cdot {{c}^{k}}).\dfrac{{{x}^{n-k}}}{{{x}^{k}}}=((\begin{matrix} n \\\ k \\\ \end{matrix})\cdot {{c}^{k}}).{{x}^{n-2k}}
Therefore, the middle term is constant in this case that is k=n2k=\dfrac{n}{2}.

So, the constant term in a binomial expression which is (x+y)n{{(x+y)}^{n}} is yn{{y}^{n}}.

Note:
For the above solution, there was one more case but it has no constant term. The binomial expression for the third case is (x+y)n{{(x+y)}^{n}}. We will study the details of the binomial theorem in the coming lessons.