binomial expansion question (1 Viewer)

saltshaker · Sep 29, 2021

Screen Shot 2021-09-29 at 2.53.49 pm.png

(cambridge 15C)

answer: 40C20

I attempted this by first seeing that every term of (x+1/x)^40 multiplied by its reciprocal from (x-1/x)^40 creates a constant and is also cancelled out

e.g. x^40 * -x^-40 = -1, x^-40 * x^40 = 1, cancels out because of symmetry

However the remaining middle terms don't multiply to 40C20...

Drongoski · Sep 29, 2021

CM_Tutor · Sep 29, 2021

saltshaker said:
However the remaining middle terms don't multiply to 40C20...

I'm going to guess that they do multiply to 40C20.

A useful identity:

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

and I am guessing that you got the LHS of this, with $n = 20$ .

@Drongoski has shown you how to get the RHS and your expansion approach gives the LHS of the specific $n = 20$ case of this identity:

$\binom{20}{0}^2 + \binom{20}{1}^2 + \binom{20}{2}^2 + ... + \binom{20}{20}^2 = \binom{40}{20}$

saltshaker · Sep 29, 2021

CM_Tutor said:
I'm going to guess that they do multiply to 40C20.

A useful identity:

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

and I am guessing that you got the LHS of this, with $n = 20$ .

@Drongoski has shown you how to get the RHS and your expansion approach gives the LHS of the specific $n = 20$ case of this identity:

$\binom{20}{0}^2 + \binom{20}{1}^2 + \binom{20}{2}^2 + ... + \binom{20}{20}^2 = \binom{40}{20}$

if i decided to use that in an exam would i have to prove it first?

CM_Tutor · Sep 29, 2021

saltshaker said:
if i decided to use that in an exam would i have to prove it first?

I think it is unlikely to appear without some indication that you need to prove it.

But, if you really needed it and a proof was not being requested, you could sketch one, like:

$\begin{align*} \text{Consider the identity}\ (1 + x)^{2n} &= x^n(1 + x)^n\left(1 + \cfrac{1}{x}\right)^n \\ \text{Term in $x^n$ on LHS}\ &= \binom{2n}{n}x^n \\ \text{Term in $x^n$ on RHS}\ &= x^n\left[\binom{n}{0} \times \binom{n}{0} + \binom{n}{1}x \times \binom{n}{1}\cfrac{1}{x} + \binom{n}{2}x^2 \times \binom{n}{2}\cfrac{1}{x^2} + ... + \binom{n}{n}x^n \times \binom{n}{n}\cfrac{1}{x^n}\right] \\ &= \left[\binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + ... + \binom{n}{n}^2\right]x^n \\ \text{It follows, by equating coefficients, that} \quad \binom{2n}{n} &= \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + ... + \binom{n}{n}^2 \end{align*}$

CM_Tutor · Sep 29, 2021

A problem in which it would come up, though in an exam it would have several parts, would be:

Bill and Ted are bored in lockdown, and so decide to play a game. Each of them tosses a fair coin fifty times. Show that the probability that they each toss the same number of heads is $\cfrac{100!}{2^{100}\left(50!\right)^2$ .

binomial expansion question (1 Viewer)

saltshaker

Member

Drongoski

Well-Known Member

CM_Tutor

Moderator

saltshaker

Member

CM_Tutor

Moderator

CM_Tutor

Moderator

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