Content
Applying the binomial theorem
In this section, we give some examples of applying the binomial theorem.
Example
Expand (2x+3)4.
Solution
Using the binomial theorem:
(a+b)4=a4+4a3b+6a2b2+4ab3+b4.Let a=2x and b=3. Then
(2x+3)4=(2x)4+4(2x)3×3+6(2x)2×32+4(2x)×33+34=16x4+96x3+216x2+216x+81.Example
Expand (1−x)10.
Solution
Using the binomial theorem:
\begin{align*} (a+b)^n &= \sum_{r=0}^{n}\dbinom{n}{r}a^{n-r}b^r. \end{align*}Let a=1, b=-x and n = 10. Then
\begin{align*} &(1-x)^{10} = \sum_{r=0}^{10}\dbinom{10}{r}(-x)^r\\ &= 1-10x+45x^2-120x^3+210x^4-252x^5+210x^6-120x^7+45x^8-10x^9+x^{10}. \end{align*}The general term in the binomial expansion
The general term in the expansion of (a+b)^n is
\dbinom{n}{r}a^{n-r}b^r, \qquad\text{where}\quad 0 \leq r\leq n.Example
For the expansion of (x-3z^2)^5, find the general term and find the coefficient of x^3z^4.
Solution
The general term is
\dbinom{5}{r}x^{5-r}(-3z^2)^r, \qquad\text{where}\quad 0 \leq r\leq 5.To find the coefficient of x^3z^4, we want 5-r=3 and r=2.
Thus the coefficient of x^3z^4 is \dbinom{5}{2}(-3)^2 = 90.
Exercise 6
Consider the expression (a-2b^3)^7. How many terms are there in the expansion? Find a formula for the general term.
Example
Find the constant term in the expansion of \Big(x^2 + \dfrac{1}{x^2}\Big)^6.
Solution
The general term is
\begin{align*} \dbinom{6}{r}\big(x^2\big)^{6-r}\Big(\dfrac{1}{x^2}\Big)^r &= \dbinom{6}{r}x^{12-2r}\Big(\dfrac{1}{x^{2r}}\Big)\\ &=\dbinom{6}{r}x^{12-4r}. \end{align*}This term will be a constant when 12-4r=0, that is, when r=3.
Hence the constant term in the expansion of \Big(x^2 + \dfrac{1}{x^2}\Big)^6 is \dbinom{6}{3}= 20.
Example
Find the coefficient of x^5 in the expansion of (1-2x+3x^2)^5.
Solution
Bracket 1-2x and expand:
(1-2x+3x^2)^5=(1-2x)^5+5(1-2x)^4(3x^2)+10(1-2x)^3(3x^2)^2+\dotsb.- The coefficient of x^5 in (1-2x)^5 is (-2)^5 = -32.
- The coefficient of x^5 in 5(1-2x)^4(3x^2) is 15\tbinom{4}{3}(-2)^3 = -480.
- The coefficient of x^5 in 10(1-2x)^3(3x^2)^2 is 90\tbinom{3}{1}(-2) = -540.
- The remaining terms do not contain x^5.
Therefore the required coefficient is -32-480-540=-1052.
The middle term
When n is even, there will be an odd number of terms in the expansion of (a+b)^n, and hence there will be a middle term. Let n=2m, for some positive integer m. Then, when the expansion of (a+b)^n is arranged with terms in descending or ascending order, the middle term is
\dbinom{2m}{m}a^mb^m.For example, the middle term of (a+b)^8 is
\dbinom{8}{4}a^4b^4 = 70a^4b^4.When n is odd, there will be two middle terms of (a+b)^n. Let n=2m+1, for some positive integer m. Then the middle terms are
\dbinom{2m+1}{m}a^{m+1}b^m \qquad\text{and}\qquad \dbinom{2m+1}{m+1}a^mb^{m+1}.Greatest coefficients
When n is even, we can show that the greatest coefficient of (1+x)^n is the
coefficient of the middle term, which is
\dbinom{n}{\dfrac{n}{2}}.When n is odd, there are two greatest coefficients, which are the coefficients of the
two middle terms
\dbinom{n}{\dfrac{1}{2}(n-1)} \qquad\text{and}\qquad \dbinom{n}{\dfrac{1}{2}(n+1)}.In general, finding the greatest coefficients for an expansion can be undertaken in a
systematic fashion, as shown in the following example.
Example
Find the greatest coefficient in the expansion of (1+ 3x)^{21}.
Solution
The general term of this expansion is \dbinom{21}{k}(3x)^k, and its coefficient is c_k = \dbinom{21}{k}3^k. The next coefficient is c_{k+1}= \dbinom{21}{k+1}3^{k+1}.
We have
\begin{align*} \dfrac{c_{k+1}}{c_k}&=\dfrac{\tbinom{21}{k+1}3^{k+1}}{\tbinom{21}{k}3^k}\\ &=\dfrac{21!}{(20-k)!(k+1)!} \times \dfrac{(21-k)!k!}{21!} \times 3\\ &=\dfrac{63-3k}{k+1}. \end{align*}To find where the coefficients are increasing, we solve c_{k+1}>c_k, that is,
\dfrac{c_{k+1}}{c_k} >1.
From above, \dfrac{63-3k}{k+1}>1, which is
equivalent to k<15\dfrac{1}{2}.
Now k is an integer, and hence c_{k+1}>c_k for k=0,1,2,\dots,14,15.
The sequence of coefficients is increasing from c_0 to c_{16} and decreasing from c_{16} to c_{21}.
Hence c_{16} is the largest coefficient:
c_{16}=\dbinom{21}{16}3^{16} = 875\, 957\, 725\, 629.Next page - Content - Proof of the binomial theorem by mathematical induction