STEP 2003 Solutions Thread Watch

toasted-lion
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(Original post by DeanK22)
III 5 is basically I 3 2006 with a tincy bit at the end;
I thought it seemed familiar! How strange that they repeated a III question on I.
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SimonM
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STEP II, Question 12

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\displaystyle \mathbb{P} ( X_1 = 2 \& X_2 = 2 ) = p^2 \left ( \left (\binom{2}{1}pq \right)^2 + 2 \cdot \left ( \binom{2}{0} q^2 \right ) \cdot \left ( \binom{2}{2} p^2\right ) \right ) = 6p^4q^2

\displaystyle \mathbb{P} ( X_1 = 2 | X_2 = ) = \frac{6p^4q^2}{6p^4q^2 + 2p^3q} = \frac{3pq}{3pq+1} = \frac{9}{25}

Therefore

\displaystyle 16pq =3 \Rightarrow \boxed{p \in \left \{ \frac{1}{4}, \frac{3}{4}\right \} }

\mathbb{E}(X_n) = 2p\mathbb{E}(X_{n-1}) since each of the remaining \mathbb{E}(X_{n-1} will spit out 2p particles.

Therefore \mathbb{E}(X_n) = (2p)^n

Therefore \displaystyle \lim_{n \to \infty} \mathbb{E}(X_n) \to 0 if \displaystyle p = \frac{1}{4} and \displaystyle \lim_{n \to \infty} \mathbb{E}(X_n) \to \infty if \displaystyle p = \frac{3}{4}
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SimonM
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STEP II, Question 13

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The sum over all probabilities will be 1, therefore we have:

\displaystyle 1 = Ae^{-\lambda}\sum_{k=1}^{\infty} \frac{\lambda^k}{k!} = Ae^{-\lambda}(e^{\lambda} - 1) \Rightarrow A = (1-e^{-\lambda})^{-1}

Consider the probability generating function

\displaystyle G(x) = \sum_{k=1}^{\infty} Ae^{-\lambda} \frac{\lambda^k}{k!} x^k = Ae^{-\lambda} (e^{\lambda x}-1)



Therefore \mu = G'(1) = \frac{\lambda}{e^{\lambda}-1} e^{\lambda}

And

\displaystyle Var(X) = (xG(x))'_{x=1} - \mu^2 = \mu(1-\mu + \lambda)

Since Var(X) > 0 \Rightarrow \mu < 1+ \lambda

And the formula for \mu \Rightarrow \mu > \lambda
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maltodextrin
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(Original post by SimonM)
STEP III, Question 6

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A trivial consequence of the product to sum formulae

Let S be the sum

2 \sin \frac{1}{2} \theta S = \left ( \cos \left( a + \frac{1}{2}\right ) \theta - \cos \left ( a - \frac{1}{2}\right ) \right ) + \left ( \cos \left( a +1+ \frac{1}{2}\right ) \theta - \cos \left ( a 

+1- \frac{1}{2}\right ) \right ) + \cdots + \left ( \cos \left( b-1+ \frac{1}{2}\right ) \theta - \cos \left ( b - 1-\frac{1}{2}\right ) \right ) =\cos \left( b- \frac{1}{2}\right ) - \cos \left( a- \frac{1}{2}\right )

So

S = \dfrac{\cos \left( b- \frac{1}{2}\right )\theta - \cos \left( a- \frac{1}{2}\right )\theta}{2 \sin \frac{1}{2} \theta} = -\dfrac{\sin \left ( \frac{(b+a-1)\theta}{2}\right ) \cos \left ( \frac{(b-a)\theta}{2} \right )}{\sin \frac{1}{2} \theta} = 0

Therefore (b-a)\theta = (2r+1)\pi or (b+a-1)\theta = 2r \pi
I think you've used the wrong trig identity in the second part, it should be sin(a + 1/2)x - sin(a - 1/2)x + .... sin(b - 1 + 1/2)x, which gives the answer as
x(a + b - 1) = pi(2n + 1) or x(b - a) = 2npi
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Rocious
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STEP II Question 11

First Part:

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v_x = u\cos\alpha \, \, \,

v_y = u\sin\alpha - gt

s_y = u(\sin\alpha) t - \frac{1}{2}gt^2

----------------------------------------

T_1 = t \, \, \, \text{when} \, \, \, s_y = 0

\implies u(\sin\alpha) T_1 - \frac{1}{2} gT_1^2 = 0 \implies T_1 = \frac{2u \sin\alpha}{g}

Let w = the time after projection at which the direction of motion makes an angle of 45 degrees with the horizontal.

w = t \, \, \, \text{when} \, \, \, \frac{v_y}{v_x} = \tan 45 = 1

\implies \frac{u\sin\alpha - gw}{u\cos\alpha} = 1 \implies w = \frac{u(\sin\alpha - \cos \alpha)}{g}

----------------------------------------

\frac{w}{\frac{1}{2}(1 - \cot \alpha)} = \frac{\frac{u}{g}(\sin\alpha - \cos\alpha)}{\frac{1}{2}(1 - \cot \alpha)} = \frac{2u(\sin\alpha -\cos\alpha)}{g(\frac{\sin\alpha}  {\sin{\alpha}} - \frac{\cos\alpha}{\sin\alpha})} = \frac{2u\sin\alpha}{g}

\implies \frac{w}{\frac{1}{2}(1 - \cot\alpha)} = T_1 \implies w = \frac{1}{2}(1 - \cot\alpha)T_1


Second Part:

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Okay so the situation at t = \frac{u}{g}(\sin\alpha - \cos\alpha) is this:

P_2's speed along y is m.
P_2 is h above the ground.

m = u\sin\alpha - g[\frac{u}{g}(\sin\alpha - \cos\alpha)] = u\cos\alpha

and then that doubles to 2u\cos\alpha.

h = u\sin\alpha[\frac{u}{g}(\sin\alpha - \cos\alpha)] - \frac{1}{2}g[\frac{u}{g}(\sin\alpha - \cos\alpha)]^2 = \frac{u^2}{2g}(\sin^2\alpha - \cos^2\alpha)

so

s_{y(ii)} = h + 2mt - \frac{1}{2}gt^2 = \frac{u^2}{2g}(\sin^2\alpha - \cos^2\alpha) + 2u(\cos\alpha) t - \frac{1}{2}gt^2

so

T_2 = w + [ t \, \, \, \text{when} \, \,\,s_{y(ii)} = 0 ]

----------------------------------------

\frac{u^2}{2g}(\sin^2\alpha - \cos^2\alpha) + 2u\cos\alpha t - \frac{1}{2}gt^2 = 0

\implies gt^2  - 4u(\cos\alpha)t  - \frac{u^2}{g}(\sin^2\alpha - \cos^2\alpha) = 0*

\implies t = \frac{4u\cos\alpha + \sqrt{16u^2\cos^2\alpha + 4u^2(\sin^2\alpha - \cos^2\alpha)}}{2g} = \frac{2u\cos\alpha + u\sqrt{3\cos^2\alpha + \sin^2\alpha}}{g}

16u^2\cos^2\alpha + 4u^2(\sin^2\alpha - \cos^2\alpha) = (4u\cos\alpha)^2 - 4u^2\cos 2\alpha

cos2a < 0 because 90 < 2a < 180

so \sqrt{(4u\cos\alpha)^2 - 4u^2\cos 2\alpha} &gt; 4u\cos\alpha and time can't be negative so yeah, the second root can be disregarded.

----------------------------------------

T_2 = \frac{2u\cos\alpha + u\sqrt{3\cos^2\alpha + \sin^2\alpha}}{g} + \frac{u(\sin\alpha - \cos\alpha)}{g} = \frac{u\sin\alpha + u\cos\alpha + u\sqrt{3\cos^2\alpha + \sin^2\alpha}}{g}

\frac{2T_2}{T_1} = \left(\frac{2u(\sin\alpha + \cos\alpha + \sqrt{3\cos^2\alpha + \sin^2\alpha})}{g}\right)\left(\  frac{g}{2u\sin \alpha }\right)

=\frac{\sin\alpha}{\sin\alpha} + \frac{\cos\alpha}{\sin\alpha} + \frac{\sqrt{3\cos^2\alpha + \sin^2\alpha}}{\sin\alpha}

= 1 + \cot \alpha + \sqrt{\frac{3\cos^2\alpha + \sin^2\alpha}{\sin^2\alpha}} = 1 + \cot \alpha + \sqrt{3\cot^2\alpha + 1}

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Elongar
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STEP II, Question 9

Solution One:

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This is the algebra cesspool...

Let the reaction forces on the rod at A and C be R_1 and R_2 respectively, so that the friction forces at these points are R_1 \tan \lambda_1 and R_2 \tan \lambda_2.

Resolving horizontally gives:

\displaystyle R_1 \tan \lambda_1 + R_2 \tan \lambda_2 \cos \alpha = R_2 \sin \alpha ~~~~~~~~~~~~~~~~~~~~(1)

Resolving vertically:

\displaystyle W = R_1 + R_2 \cos \alpha + R_2 \tan \lambda_2 \sin \alpha ~~~~~~~~~~~~~~~~~~~~~~(2)

From (1), rearranging gives us R_1 in terms of R_2:

\displaystyle \displaystyle R_1 = \frac{R_2 \sin \alpha - R_2 \tan \lambda_2 \cos \alpha}{\tan \lambda_1}

Substituting this for R_1 in (2), multiplying by \sin \lambda_1 gives:

\displaystyle W \sin \lambda_1 = (R_2 \sin \alpha \cos \lambda_1 + R_2 \cos \alpha \sin \lambda_1) - (R_2 \tan \lambda_2 \cos \alpha \cos \lambda_1 - R_2 \tan \lambda_2 \sin \alpha \sin \lambda_1)

Using compound-angle formulae reduces this to:

\displaystyle W \sin \lambda_1 = R_2 \sin (\alpha + \lambda_1) - R_2 \tan \lambda_2 \cos (\alpha + \lambda_1)

Rearranging for R_2:

\displaystyle R_2 = \frac{W \sin \lambda_1 \cos \lambda_2}{\sin (\alpha + \lambda_1 - \lambda_2)}

Finally, substituting this back into (2) and rearranging for R_1 gives:

\displaystyle R_1 = \frac{W \cos \lambda_1 \sin (\alpha - \lambda_2)}{\sin (\alpha + \lambda_1 - \lambda_2)}

We want the magnitude of the friction force, R_1 \tan \lambda_1, which is therefore:

\displaystyle F_1 = \frac{W \sin \lambda_1 \sin (\alpha - \lambda_2)}{\sin (\alpha + \lambda_1 - \lambda_2)}

QED



Solution Two:

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Significantly cleaner.

As before, let the reaction forces on the rod at A and C be R_1 and R_2 respectively, so that the friction forces at these points are R_1 \tan \lambda_1 and R_2 \tan \lambda_2.

\lambda_1 and \lambda_2 are the so-called "angles of friction", and describe the angle between the reaction forces and the friction forces.

Since our two friction forces and the weight of the rod result in an equilibrium, we can draw a force triangle.

Then:

\angle WC_1 = \lambda_1

\angle WC_2 = \alpha - \lambda_2

and

\angle C_1C_2 = \pi - (\alpha + \lambda_1 - \lambda_2)

Simply applying the sine rule leads to the result.
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Mark13
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(Original post by Dadeyemi)
STEP II 2003 Q5
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E = a/2, F = (p+b)/2, G = b/2, H = (p+a)/2 =>
EF = a/2+x[p+b-a]/2
GH= b/2+y[p+a-b]/2
for which there is a solution EF = GH = (p+a+b)/4 when x = y = 1/2
so we have s = (d+t)/2 = (p+a+b)/4 so the position vector of T is t = (p+a+b)/4-d/2
the plane OAB is the x-y plane i.e. z = 0 so the component of T in the k direction is 0, so considering the vectors in the k direction we have: 0 = n/4-d/4 <=> d = 2n
I'm fairly sure the answer is
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d=n/2

Consider that the z co-ordinate of H is the average of the z co-ordinates of A (0) and P (n). So H has a a z co-ordinate of n/2. Similarly, S, the midpoint of AH, has a z co-ordinate of n/4. If S is the midpoint between D and T (which we are told has a z co-ordinate of 0), then D must have a z co-ordinate of n/2. Therefore d=n/2
.
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Dadeyemi
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(Original post by Mark13)
I'm fairly sure the answer is
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d=n/2

Consider that the z co-ordinate of H is the average of the z co-ordinates of A (0) and P (n). So H has a a z co-ordinate of n/2. Similarly, S, the midpoint of AH, has a z co-ordinate of n/4. If S is the midpoint between D and T (which we are told has a z co-ordinate of 0), then D must have a z co-ordinate of n/2. Therefore d=n/2
.
Yea I just made some rediculously silly mistakes (e.g. forgetting to multiply by two) I corrected it and get the same answer.
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around
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(Original post by maltodextrin)
I think you've used the wrong trig identity in the second part, it should be sin(a + 1/2)x - sin(a - 1/2)x + .... sin(b - 1 + 1/2)x, which gives the answer as
x(a + b - 1) = pi(2n + 1) or x(b - a) = 2npi
I did using 'otherwise' and I got that answer as well.
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maltodextrin
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(Original post by tommm)
STEP III 2003, Q4

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\frac{dy}{dt} = 1 + 3t^2, \frac{dx}{dt} = 2t, therefore \frac{dy}{dx} = \frac{1 + 3t^2}{2t}

Using y - y_1 = m(x - x_1), we obtain the equation of the tangent:

y - t(1 + t^2) = \frac{1 + 3t^2}{2t}(x - t^2)

This meets the curve again when x = T^2, y = T(1 + T^2). Substituting this in and rearranging to find a quadratic in T^2, we obtain:

(2t)T^3 - (1 + 3t^2)T^2 + (2t)T + (t^4 - t^2) = 0

By the factor theorem, we find that T = t is a root. We can factorise and divide by (T - t) (because T \neq t) to obtain

(2t)T^2 - (1 + t^2)T + (t - t^3) = .

Using the quadratic formula, we obtain

T = \displaystyle\frac{1 + t^2 \pm (1 - 3t^2)}{4t}

One of these gives T = t, which we can disregard.

Therefore, T = \frac{1 - t^2}{2t} as required.



The double angle formula for tangent is:

\tan 2\theta = \displaystyle\frac{2\tan \theta}{1 - \tan^2 \theta}

Therefore, \cot 2\theta = \displaystyle\frac{1 - \tan^2 \theta}{2\tan \theta}

So, if t_i = \tan \theta, then t_{i + 1} = \cot 2\theta

Therefore, if t_0 = \tan (\frac{7\pi}{18}), then
t_1 = \cot (\frac{7\pi}{9})

We can also apply the formula similarly to the cotangent function, so

t_2 = \tan (\frac{14\pi}{9})
t_3 = \cot (\frac{28\pi}{9})

Using the identities \cot(\theta + \pi) = \cot \theta and \cot(\frac{\pi}{2} - \theta) = \tan \theta, we get

t_3 = \tan (\frac{7\pi}{18}) = t_0 as required.

Another value of t_0 which satisfies the required properties is \tan(\frac{\pi}{9}).

I might be misinterpreting the question but the question asks us to 'show that T = (1 - t^2)/2t and 3t^2 =/ 1'. So this means we should give some justification as to why 3t^2 =/1 right? If so then what would the justification be?

If 3t^2 = 1 then it means that T = (1 + t^2)/4t which I can't see a problem with...
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Mark13
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(Original post by maltodextrin)
I might be misinterpreting the question but the question asks us to 'show that T = (1 - t^2)/2t and 3t^2 =/ 1'. So this means we should give some justification as to why 3t^2 =/1 right? If so then what would the justification be?

If 3t^2 = 1 then it means that T = (1 + t^2)/4t which I can't see a problem with...
I think it's because you need T=t to be a repeated root, since its a tangent. Not sure though.
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Mark13
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(Original post by SimonM)
STEP III, Question 6

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A trivial consequence of the product to sum formulae

Let S be the sum

2 \sin \frac{1}{2} \theta S = \left ( \cos \left( a + \frac{1}{2}\right ) \theta - \cos \left ( a - \frac{1}{2}\right ) \right ) + \left ( \cos \left( a +1+ \frac{1}{2}\right ) \theta - \cos \left ( a 

+1- \frac{1}{2}\right ) \right ) + \cdots + \left ( \cos \left( b-1+ \frac{1}{2}\right ) \theta - \cos \left ( b - 1-\frac{1}{2}\right ) \right ) =\cos \left( b- \frac{1}{2}\right ) - \cos \left( a- \frac{1}{2}\right )

So

S = \dfrac{\cos \left( b- \frac{1}{2}\right )\theta - \cos \left( a- \frac{1}{2}\right )\theta}{2 \sin \frac{1}{2} \theta} = -\dfrac{\sin \left ( \frac{(b+a-1)\theta}{2}\right ) \cos \left ( \frac{(b-a)\theta}{2} \right )}{\sin \frac{1}{2} \theta} = 0

Therefore (b-a)\theta = (2r+1)\pi or (b+a-1)\theta = 2r \pi
Don't think this is right; there shouldn't be any multiples of 2\pi in the solution set.
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maltodextrin
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(Original post by Mark13)
Don't think this is right; there shouldn't be any multiples of 2\pi in the solution set.
Yeah I think SimonM just used the wrong trig identity. I think I posted the correct version somewhere in the thread.
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Evan247
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I'd like to ask, should i figure out whether f(1) or f(2) is greater in each case?
(Original post by tommm)
STEP II 2003 Q8

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\frac{dy}{dt} = -ky(\displaystyle\frac{t^2 - 3t + 2}{t + 1})

\implies \displaystyle\int\frac{\mathrm{d  }y}{y} = -k\displaystyle\int\displaystyle\  frac{t^2 - 3t + 2}{t + 1}\mathrm{d}t

\implies \ln y = -k\displaystyle\int(t - 4 + \frac{6}{t+1})\mathrm{d}t = -k(\frac{1}{2}t^2 - 4t + 6\ln(t + 1) + c)

\implies y = Ae^{-k(\frac{1}{2}t^2 - 4t + \ln((t + 1)^6)}

\implies y = A(t + 1)^{-6k}e^{-k(\frac{1}{2}t^2 - 4t)}


Stationary values occur when \frac{dy}{dt} = 0. We will use this in the original differential equation, so we obtain

y(\displaystyle\frac{t^2 - 3t + 2}{t + 1} = 0

\implies y(t-1)(t-2) = 0, t \neq -1

Now, if y = 0, then from our expression for y, t = -1, which would lead to a fraction in our equation for \frac{dy}{dx} = 0 becoming infinite. Therefore, we disregard y = 0 as a stationary point.
Therefore, stationary points occur at t = 1, 2

So y(1) = A \times 2^{-6k}e^{-3.5k}, y(2) = A \times 3^{-6k}e^{-6k}

Therefore \frac{y(2)}{y(1)} = (3/2)^{-6k}e^{-\frac{5}{2}k} as required.

When k &gt; 0, y \rightarrow +0 as t \rightarrow +\infty

When k &lt; 0, y \rightarrow +\infty as t \rightarrow +\infty

I've used a graphing program to draw the two graphs. The first is in the case A = 1, k = 1; the second is when A = 1, k = -1. Note the location of the turning points.







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Evan247
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I got same solutions as yours
(Original post by toasted-lion)
Hmm... yuk! Consider the area of the sector OGH and the triangle OGH, subtracting to get the segment. Nice perseverance, but it's actually pretty easy.


Are you sure we want the distance to the edge and not to the centre? We had d as the distance to centre before. I got:

for (i)  d = |c-b| (note the x-coord of the centre is irrelevant)
for (ii)  d = \frac{|c|}{\sqrt{m^2 +1}} (perpendicular distance)
for (iii)  d = \frac{|ma + c - b|}{\sqrt{m^2 +1}} (perpendicular distance again)

and I just plugged these straight into the equation from the first part. As you noted (i) and (ii) follow from (iii), but I'm sure they put them in that order for a reason (ie to give you bite-sized chunks of the overall thought-process). Please tell me if I made a mistake

P.S. You got a stray /latex tag :P
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nota bene
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(Original post by SimonM)
III/2
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\displaystyle \sum_{r=0}^{\infty} \binom{2r}{r} x^r  = (1-4x)^{-1/2}
What lead you to think this identity was useful?

When I attempted this I saw that (1-p)^(-0.5) wasn't expanding to quite 2rCr, but didn't think it was important to fix. Then went on to using p=1/5 which lead me to ^{2r}C_r (\frac{31}{125})^r as the general term, and gave up after trying other possibilities for a while.
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DFranklin
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(Original post by nota bene)
...
If you expand (1-p)^(-0.5), you should find you get 2rCr but with some extra powers of \frac{1}{2} lying around. Which you can fix by considering x = 4p.

(I'm not totally sure this is what you're asking, but still).
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nota bene
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(Original post by DFranklin)
If you expand (1-p)^(-0.5), you should find you get 2rCr but with some extra powers of \frac{1}{2} lying around. Which you can fix by considering x = 4p.

(I'm not totally sure this is what you're asking, but still).
No, I know how to fix it. My question was more supposed to be; how do you realise it is necessary to fix?

To me it isn't exactly natural to make that fix and then differentiate (aha, works!). Maybe I'm just bad at seeing things... Making that fix would be simple if I knew I was working towards something seemigly requiring it, but the already differentiated version where p=1/5 has been put in doesn't scream that similar to the series just found in i) although there is obviously some connection.
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DFranklin
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I think the first part is supposed to actually lead you in the right direction. That is, the first part is trying to get you to realise (1-4p)^{-1/2} = \sum \binom{2n}{n} p^n.

From there, it's fairly easy to fiddle with that identity to get something you can differentiate to get (ii).

[As a general comment, in a STEP-type exam, if you've just found a sum f(x) = \sum a_n x^n (for particular a_n), and you're asked for a sum \sum a_n P(n) x^n, where P(n) is a small polynomial in n, it is very likely that you will need to differentiate the first series (or some simple variant of the first series) to get something looking like the 2nd. Similarly, if you're asked for \sum \frac{a_n x^n}{n+k} you should be thinking about integration. And in general, the last part of a STEP question usually has some connection to the earlier parts. So I guess I'm saying (ii) should scream out as similar to (i), at least to the degree of thinking "there must be a way of manipulating one of these series to end up with the other one"].

I'm still not sure if that is what you're actually asking. I'd be surprised if any of this was particularly new to you.
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nota bene
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(Original post by DFranklin)
I think the first part is supposed to actually lead you in the right direction. That is, the first part is trying to get you to realise (1-4p)^{-1/2} = \sum \binom{2n}{n} p^n.
I didn't need to use that in the first part, I just substituted \frac{\binom {2n}{n}}{2^n.2^n}=\frac{2^n(2n-1)!}{2^n.2^n.n!} so didn't really make it 'nice' first, because it was easy to just multiply the thing by 1/(2^(2n)). I guess that's the reason I failed to see it was trying to hint me at anything.

I'm still not sure if that is what you're actually asking. I'd be surprised if any of this was particularly new to you.
I'm quite satisfied! No, nothing of this latter part is particularily new to me. It's just that in this case I didn't have the starting point clear and then it's diffícult to do something with it. It's the type of question I'll abandon after staring 5-10 minutes at it in a real exam.
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