Binomial expansion (1 Viewer)

NexusRich · Jun 30, 2021

Can someone please explain how to do this type of questions, and also what is the big idea/ purpose of 'equating the coefficients' ? Thanks in advance

tickboom · Jun 30, 2021

This is eerily similar to Q14 a) of the 2020 Extension 1 exam. I made a video working through the explanation, which you might find helpful when thinking in general about the benefit of the "equating of coefficients" technique.

Enjoy!

CM_Tutor · Jun 30, 2021

This problem has come up om quite a few exams in different forms, including as part of larger questions, like:

1. Show this result.

2. A and B each toss a fair coin $n$ times. If one tosses more heads than the other, s/he wins... if it's a draw, they toss again.
(a) Find the probability that they each toss the same number of heads.
(b) Hence, show that the probability of A winning on the first try is $\cfrac{1}{2} - \cfrac{(2n)!}{2^{2n+1}(n!)^2$ .

CM_Tutor · Jun 30, 2021

NexusRich said:
Can someone please explain how to do this type of questions, and also what is the big idea/ purpose of 'equating the coefficients' ? Thanks in advance

Equating coefficients is a useful technique for proving identities by making two equivalent forms of the same expression.

For example, suppose I wanted to write

$2x^2 - 7x + 5$

in the form

$A(x + 1)^2 + B(x + 1) + C$ .

I can expand the second form and equate coefficients to find the values of $A,\ B,\ \text{and}\ C$ that make the identity

$2x^2 - 7x + 5 \equiv A(x + 1)^2 + B(x + 1) + C$

true for all values of $x$ . I would find that $A = 2$ , that $B = -11$ and that $C = 14$ , and thus that the identity is

$2x^2 - 7x + 5 \equiv 2(x + 1)^2 - 11(x + 1) + 14$

Why would this be something I might use, you may well ask... Well, suppose I sought $\int \cfrac{2x^2 - 7x + 5}{x + 1}\ dx$ .

Using the identity, I can find the integral:

$\begin{align*} \int \cfrac{2x^2 - 7x + 5}{x + 1}\ dx &= \int \cfrac{2(x + 1)^2 - 11(x +1) + 14}{x + 1}\ dx \\ &= \int \cfrac{2(x + 1)^2}{x + 1} - \cfrac{11(x +1)}{x + 1} + \cfrac{14}{x + 1}\ dx \\ &= \int 2(x + 1) - 11 + \cfrac{14}{x + 1}\ dx \\ &= \int 2x - 9 + \cfrac{14}{x + 1}\ dx \\ &= \cfrac{2x^2}{2} - 9x + 14\ln{|x + 1|} + C,\ \text{for some constant $C$} \\ &= x^2 - 9x + 14\ln{|x + 1|} + C \end{align*}$

In the area of binomials, equating coefficients is one of the techniques used to solve problems and to prove identities.

CM_Tutor · Jul 1, 2021

In order to let you work on the solution yourself, I will offer one based on a different identity:

$\textbf{Lemma}: \qquad \qquad x^n(1+ x)^n\left(1 + \cfrac{1}{x}\right)^n = (1 + x)^{2n}$

$\begin{align*} \text{LHS}\ &= x^n(1+ x)^n\left(1 + \cfrac{1}{x}\right)^n \\ &= (1+ x)^n\left[x^n\left(1 + \cfrac{1}{x}\right)^n\right] \\ &= (1+ x)^n\left[x\left(1 + \cfrac{1}{x}\right)\right]^n \\ &= (1+ x)^n\left(x + \cfrac{x}{x}\right)^n \\ &= (1+ x)^n(x + 1)^n \\ &= \left[(1+ x)^n\right]^2 \\ &= (1 + x)^{2n} \\ &=\ \text{RHS} \end{align*}$

$\textbf{Theorem}: \qquad \qquad {\binom{n}{0}}^2 + {\binom{n}{1}}^2 + {\binom{n}{2}}^2 + ... + {\binom{n}{n}}^2 = \binom{2n}{n}$

Considering the RHS of the identity established in the above Lemma, we can expand and see that:

$\begin{align*} (1 + x)^{2n} &= \binom{2n}{0} \times 1^{2n} \times x^0 + \binom{2n}{1} \times 1^{2n-1} \times x^1 + \binom{2n}{2} \times 1^{2n - 2} \times x^2 + ... + \binom{2n}{n} \times 1^{2n - n} \times x^n + ... + \binom{2n}{2n-1} \times 1^{2n - (2n -1)} \times x^{2n - 1} + \binom{2n}{2n} \times 1^{2n - 2n} \times x^{2n} \\ &= \binom{2n}{0} + \binom{2n}{1} x + \binom{2n}{2} x^2 + ... + \binom{2n}{n} x^n + ... + \binom{2n}{2n-1} x^{2n - 1} + \binom{2n}{2n} x^{2n} \end{align*}$

It is clear that the term in $x^n$ in this expansion is $\binom{2n}{n}x^n$ . Hence, the the term we seek for our result, $\binom{2n}{n}$ , is the coefficient of $x^n$ in the expansion of $(1 + x)^{2n}$ . It is therefore reasonable to conjecture that the other part of the result we seek will be the coefficient of $x^n$ in the expansion of the LHS of the identity established in the Lemma.

Expanding the terms separately:

$(1 + x)^n = \binom{n}{0} + \binom{n}{1} x + \binom{n}{2} x^2 + ... + \binom{n}{n-1} x^{n - 1} + \binom{n}{n} x^{n}$

$\left(1 + \cfrac{1}{x}\right)^n = \binom{n}{0} + \binom{n}{1} \cfrac{1}{x} + \binom{n}{2} \cfrac{1}{x^2} + ... + \binom{n}{n-1} \cfrac{1}{x^{n - 1}} + \binom{n}{n} \cfrac{1}{x^n}$

Now, every term in the expansion of $x^n(1+ x)^n\left(1 + \cfrac{1}{x}\right)^n$ will consist of the product of $x^n$ and a term from the first expansion and a term from the second expansion. Thus:

$\begin{align*} \text{Term in $x^n$}\ &= x^n \times \binom{n}{0} \times \binom{n}{0} + x^n \times \binom{n}{1} x \times \binom{n}{1} \cfrac{1}{x} + x^n \times \binom{n}{2} x^2 \times \binom{n}{1} \cfrac{1}{x^2} + ... + x^n \times \binom{n}{n-1} x^{n-1} \times \binom{n}{n - 1} \cfrac{1}{x^{n - 1}} + x^n \times \binom{n}{n} x^n \times \binom{n}{n} \cfrac{1}{x^n} \\ &= x^n \left[{\binom{n}{0}}^2 + {\binom{n}{1}}^2 \cfrac{x}{x} + {\binom{n}{2}}^2 \cfrac{x^2}{x^2} + ... + {\binom{n}{n-1}}^2 \cfrac{x^{n-1}}{x^{n-1}} + {\binom{n}{n}}^2 \cfrac{x^n}{x^n}\right] \\ &= x^n \left[{\binom{n}{0}}^2 + {\binom{n}{1}}^2 + {\binom{n}{2}}^2 + ... + {\binom{n}{n-1}}^2 + {\binom{n}{n}}^2\right] \end{align*}$

It thus follows from equating the coefficients of $x^n$ from the LHS and RHS of the identity established in the Lemma that we have found that:

${\binom{n}{0}}^2 + {\binom{n}{1}}^2 + {\binom{n}{2}}^2 + ... + {\binom{n}{n-1}}^2 + {\binom{n}{n}}^2 = \binom{2n}{n}$

as expected.

Binomial expansion (1 Viewer)

NexusRich

Member

Attachments

tickboom

Member

CM_Tutor

Moderator

CM_Tutor

Moderator

CM_Tutor

Moderator

Users Who Are Viewing This Thread (Users: 0, Guests: 1)