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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 (1x)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*}
Exercise 5

Expand

  1. (2x-3y)^4
  2. \Big(x-\dfrac{2}{x}\Big)^4.

Screencast of exercise 5 mp4 of screencast of exercise 5

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.

Exercise 7

Find the constant term in the expansion of \Big(x + \dfrac{1}{x^2}\Big)^6.

Screencast of exercise 7 mp4 of screencast of exercise 7

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.

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}.

Exercise 8

Find the middle term of (2x-3y)^6.

Screencast of exercise 8 mp4 of screencast of exercise 8

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.
Exercise 9

Find the greatest coefficient of (2x+3y)^{15}.

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