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

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

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.