Find the coefficient of x^2 in the expansion of (x-2/x)^6

In this section, we give some examples of applying the binomial theorem.

Expand $(2x+3)^4$.

Solution

Using the binomial theorem:

\[ (a+b)^4 =a^4+4a^3b+6a^2b^2 +4ab^3+ b^4. \]

Let $a =2x$ and $b=3$. Then

\begin{align*} (2x+3)^4 &= (2x)^4+ 4(2x)^3\times3+6(2x)^2\times3^2+ 4(2x)\times3^3+ 3^4\\ &=16x^4+ 96x^3+216x^2+216x+ 81. \end{align*}

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*}

Expand

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

Screencast of exercise 5

The general term in the binomial expansion

The general term in the expansion of $(a+b)^n$ is

\[ \dbinom{n}{r}a^{n-r}b^r, \qquad\text{where}\quad 0 \leq r\leq n. \]

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 $\dbinom{5}{2}(-3)^2 = 90$.

Consider the expression $(a-2b^3)^7$. How many terms are there in the expansion? Find a formula for the general term.

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 $\dbinom{6}{3}= 20$.

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

Screencast of exercise 7

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}. $

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

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.

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:

\[ c_{16}=\dbinom{21}{16}3^{16} = 875\, 957\, 725\, 629. \]

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

Next page - Content - Proof of the binomial theorem by mathematical induction

Hint. Write the product as $$-x^{-6}(1-x)^3(1+2x^2)^6.$$ Then the general term is given by $$-x^{-6}{3\choose j}(-x)^j{6\choose k}(2x^2)^k,$$ with $0\le j\le 3,\,0\le k\le 6.$ To get the term in $x^{-2},$ you therefore need to have $-2=-6+j+2k,$ or $j+2k=4,$ whose solutions are given by $(j,k)=(0,2),\,(2,1).$ Thus you want to compute the coefficient $$(-1)^{j+1}2^k{3\choose j}{6\choose k}$$ for these two solutions and add them up.

Kristiana P.

asked • 04/30/14

In the expansion of the equation, find the coefficient of x2.

1 Expert Answer

Daniel J. answered • 05/17/14

Master's Math tutor SAT specialty

By the Binomial Theorem, [2x-(1/2x)]^6 can be expanded to the sum as k goes from zero to 6 of (6 choose k)×(2x)k×(-1/2x)6-k. The only time we will be left with an x2 term is when k=4, since we'll have some coefficient multiplied by x4/x2. So, we'll have (6 choose 4) × (2x)4×(-1/2x)2, which is 15×16x4×(1/4x2)= 60x2.

So the coefficient is 60.