STEP maths I, II, III 1990 solutions Watch

nota bene
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I just had a quick go at I/14, and have not checked my working and reasoning, so likely loads of mistakes. (Also I'm not 100% sure what the last part wants)

With Juggins strategy to draw trice and then stop there are eight possible ways of not ending up with a score of 0, WWW, RRR, WRR (in any order) and WWR (in any order).
P(WWW)=1/8
P(RRR)=(3/10)^3=27/1000
P(WRR in any order)=3\times\frac{9}{200}
P(WWR in any order)=3\times\frac{3}{40}

\frac{1}{8}+\frac{27}{1000}+\fra  c{9}{40}+\frac{27}{200}=\frac{64  }{125}
So 1-(64/125)=61/125, which is the probability he ends up with zero points in total.

The average score is 0\times\frac{61}{125}+3\frac{1}{  8}+4\frac{9}{40}+5\frac{27}{200}  +6\frac{27}{2000}=\frac{264}{125  }

Muggins with his obtained N points will have a probability of 1/2 to obtain N+1 points, probability 3/10 to obtain N+2 points and probability 1/5 to end up at zero points. Therefore the average score is \frac{1}{2}(N+1)+\frac{3}{10}(N+  2)=\frac{4}{5}N+ \frac{11}{10}

The greatest possible average final score can be obtained by applying the above formula after each try and see if the average increases, and in that case continue - until the formula suggests the average will not increase by taking part in one more draw.

Edit: As David points out in post #86 the following should probably be added at the end for a complete solution :
Given the current score is N, we know the expected score after another go is \frac{4}{5}N + \frac{11}{10}. So the expected improvement after another go is \frac{11}{10} - \frac{N}{5} = \frac{11-2N}{10}. This is positive for N <=5 and negative for N > 5. Therefore to achieve the greatest average score, you should play until the current score is greater than 5 and then stop
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insparato
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I only looked at the bit quoted! But glad you sorted it.
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Rabite
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I remember III Q8...I got something really stupid too, so I thought it was wrong.
I'll have a go during lunch break at work tomorrow or something.
Actually, maybe I shouldn't do maths at lunch break else people might think I'm really really cool.

I barely know what a moment of inertia is, so I can't help on Q11. [edit] Oopsie, too late.

[edit again] Do people really think that putting "URGENT!!" in their topic titles makes clicking on them any more of an attractive idea?
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DFranklin
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(Original post by nota bene)
I just had a quick go at I/14, and have not checked my working and reasoning, so likely loads of mistakes. (Also I'm not 100% sure what the last part wants)
Muggins with his obtained N points will have a probability of 1/2 to obtain N+1 points, probability 3/10 to obtain N+2 points and probability 1/5 to end up at zero points. Therefore the average score is \frac{1}{2}(N+1)+\frac{3}{10}(N+  2)=\frac{4}{5}N+ \frac{11}{10}

The greatest possible average final score can be obtained by applying the above formula after each try and see if the average increases, and in that case continue - until the formula suggests the average will not increase by taking part in one more draw.
I think the algorithm you're suggesting is fine, but I'm sure the examiner expected you to actually do it!

In other words: Given the current score is N, we know the expected score after another go is \frac{4}{5}N + \frac{11}{10}. So the expected improvement after another go is \frac{11}{10} - \frac{N}{5} = \frac{11-2N}{10}. This is positive for N <=5 and negative for N > 5. Therefore to achieve the greatest average score, you should play until the current score is greater than 5 and then stop (obviously stopping earlier if unfortunate enough to draw a black ball).
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nota bene
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(Original post by DFranklin)
I think the algorithm you're suggesting is fine, but I'm sure the examiner expected you to actually do it!
Argh! You're most certainly right - and I hate the wording of the question. To me 'suggest a strategy to achieve the greatest possible average score' means I shall outline the method, and not necessarily tell the player how many times he shall repeat

edit:
(Original post by Rabite)
Actually, maybe I shouldn't do maths at lunch break else people might think I'm really really cool.
You are really really cool:cool:

Do people really think that putting "URGENT!!" in their topic titles makes clicking on them any more of an attractive idea?
Probably...
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DFranklin
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(Original post by nota bene)
Argh! You're most certainly right - and I hate the wording of the question. To me 'suggest a strategy to achieve the greatest possible average score' means I shall outline the method, and not necessarily tell the player how many times he shall repeat
It's not so bad as the wording on STEPIII,Q14 this year - I'm pretty certain the examiners stuffed up (by not specifying the square should be axis-aligned), and I'm pretty sure at least one person couldn't do the question because of it in the real exam (+SsEe here).

(Fortunately, the person who decided it was too difficult without assuming axis-alignment is the top performer on the BMO team, so I suspect it won't have affected his chances for entrance).
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ukgea
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I/6

Let \mathbf a, \mathbf b, \mathbf c, \mathbf d be the position vectors of A, B, C, D respectively. Without loss of generality, let \mathbf b = \mathbf 0. Then \mathbf d = \mathbf a + \mathbf c.

(i) The statement is equivalent to

\mathbf a^2 + \mathbf c^2 + (\mathbf a + \mathbf c - \mathbf c)^2 + (\mathbf a + \mathbf c - \mathbf c)^2 = (\mathbf a - \mathbf c)^2 + (\mathbf a + \mathbf c)^2

(where squaring a vector denotes taking the dot product of that vector with itself)
This is in turn equivalent to

\mathbf a^2 + \mathbf c^2 + \mathbf a^2 + \mathbf c^2 = \mathbf a^2 - 2\mathbf a\cdot \mathbf c + \mathbf a + 2\mathbf a \cdot \mathbf c + \mathbf c^2

which is evidently true. Thus the statement is proved.

(ii) |AC^2 - BD^2|

= | (\mathbf a - \mathbf c)^2 - (\mathbf a + \mathbf c)^2|

= | \mathbf a^2 - 2\mathbf a \cdot \mathbf c + \mathbf c^2 - \mathbf a^2 - 2\mathbf a \cdot \mathbf c - \mathbf c^2|

= |-4\mathbf a \cdot \mathbf c|

= |4 \mathbf a \cdot \mathbf c|

= |4 |\mathbf a| |\mathbf c| \cos \theta |

= 4 |\mathbf a| |\mathbf c| |\cos \theta|

where \theta is either the angle ABC or the angle BCD (it doesn't matter which, since the cosine of these two angles are negatives of each other and this distinction disappears when taking the modulus anyway.)

Now, the expression |\mathbf c| |\cos \theta| is the length of the projection of BC on AB. By symmetry, this is also the length of the projection of AD onto CD, and it is also the projection of BC on CD and AD on AB. Note that |\mathbf  a| in this case represents the length of AB or CD, and thus the whole expression can be seen as four times the area of the rectangle whose sides are either of BC, AD projected on either of AB, CD, together with AB or CD.

Instead, if we consider |\mathbf a| |\cos \theta| and |\mathbf c| we can similarly see that the expression is four times the area of the rectangle whose sides are either of AB, CD projected on BC, AD, together with BC or AD. And thus, the expression is four times the area of the rectangle formed by any side and the projection of any adjacent side onto that side, as required.

For the next case, the result we shall prove is:

|AB^2 - AD^2| is the area of the rectangle whose sides are one diagonal and the projection of the other diagonal onto the first diagonal.

This is because

|AB^2 - AD^2|

= |\mathbf a^2 - \mathbf (a + c - a)^2|

= |\mathbf a^2 - \mathbf c^2|

= |(\mathbf a + \mathbf c)\cdot(\mathbf a - \mathbf c)|

= |BD| |AC| |\cos \theta|

Where \theta this time is the angle between the diagonals. (Again, it doesn't matter if we take the obtuse or the acute angle, because of the modulus sign).

Now since |AC| |\cos \theta| and |BD| |\cos \theta| are the length of the projections of AC on BD and BD on AC respectively, we can see the above expression as either BD multiplied by length of projection of AC on BD, or AC multiplied by the length of the projection of BD onto AC. This thus proves the statement.
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Dystopia
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STEP III, Q4

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\cos\alpha + \cos(\alpha + 2\beta) + \cdots + \cos(\alpha + 2n\beta)

Adding the first and last terms, second and second-to-last terms, and so on, and using \cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}, we get:

2\cos(\alpha + n\beta)\left(\cos n\beta + \cos(n-2)\beta + \cdots \right) finishing with either \cos3\beta + \cos\beta or \cos2\beta + 1 depending on whether n is odd or even.

Let z = \cos\beta + i\sin\beta.

Then z^{n} + \frac{1}{z^{n}} = 2\cos n\beta

2(\cos n\beta + \cos(n-2)\beta + \cdots ) = (z^{n} + \frac{1}{z^{n}}) + (z^{n-2} + \frac{1}{z^{n-2}}) + \cdots

= \frac{z^{2n} + z^{2n-2} + \cdots + z^{2} + 1}{z^{n}} = \frac{z^{n+1}-z^{-(n+1)}}{z-z^{-1}} = \frac{\sin(n+1)\beta}{\sin\beta}

\displaystyle\Rightarrow \sum_{r=0}^{n} \cos(\alpha + 2r\beta) = \frac{\sin(n+1)\beta \cos(\alpha + n\beta)}{\sin\beta}

---

\cos\alpha + \binom{n}{1}\cos(\alpha + 2\beta) + \cdots + \binom{n}{r}\cos(\alpha + 2r\beta) + \cdots + \cos(\alpha + 2n\beta)

Note that \binom{n}{r} = \binom{n}{n-r}

Again adding together values from the opposite ends and using the sum to product formula, we get:

2\cos(\alpha + n\beta)\left(\cos n\beta + \binom{n}{1}\cos(n\beta-2) + \cdots \right)

z = \cos\beta + i\sin\beta

(z + \frac{1}{z})^{n} = z^{n} + \binom{n}{1}z^{n-2} + \cdots + \binom{n}{n-2}\frac{1}{z^{n-2}} + \frac{1}{z^{n}} = 2^{n}\cos^{n}\beta

\displaystyle\Rightarrow \sum_{r=0}^{n} \binom{n}{r} \cos(\alpha + 2r\beta) = 2^{n}\cos(\alpha + n\beta)\cos^{n}\beta

---

Suppose \displaystyle\sum_{r=0}^{k} \cos r\theta \sec^{r}\theta = \frac{\sin(k+1)\theta \sec^{k}\theta}{\sin\theta}

Then \displaystyle\sum_{r=0}^{k+1} \cos r\theta \sec^{r}\theta = \sec^{k}\theta\left( \frac{\sin(k+1)\theta}{\sin\thet  a} + \frac{\cos(k+1)\theta}{\cos\thet  a} \right)

 = \sec^{k}\theta \left(\frac{\sin(k+1)\theta \cos\theta + \sin\theta\cos(k+1)\theta}{\sin\  theta\cos\theta} \right)

 = \frac{\sin((k+1)+1)\theta \sec^{k+1}\theta}{\sin\theta}

The case n=0 gives 1 = \frac{\sin\theta \sec^{0}\theta}{\sin\theta}

Therefore the truth of n=k implies the truth of n=k+1, and it is also true for n=0, so the summation is true for all natural n.

---

Alternate Solution to the first part.

Suppose \cos\alpha + \cos(\alpha + 2\beta) + \cdots + \cos(\alpha + 2n\beta) = N

Then \sin\beta(\cos\alpha + \cdots + \cos(\alpha+2n\beta)) = N\sin\beta

Let \gamma = \frac{\pi}{2} + \alpha

-2\sin\beta(\sin\gamma + \cdots + \sin(\gamma + 2n\beta)) = -2N\sin\beta

Using -2\sin A \sin B = \cos(A+B) - \cos(A-B), we get:

\left(\cos(\gamma + \beta) - \cos(\gamma - \beta)\right)

+ \left(\cos(\gamma + 3\beta) - \cos(\beta + \gamma)\right)

\cdots

+ \left(\cos(\gamma + (2n+1)\beta) - \cos(\gamma + (2n-1)\beta)\right)

This acts as a 'telescoping' series, and the only terms remaining are:

\cos(\gamma + (2n+1)\beta) - \cos(\gamma - \beta) = -2\sin(n+1)\beta \sin(\gamma + n\beta)

= -2\sin(n+1)\beta\cos(\alpha + n\beta) = -2N\sin\beta

\Rightarrow N = \frac{\sin(n+1)\beta\cos(\alpha+  n\beta)}{\sin\beta}
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nota bene
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I just scribbled through an answer to I/16, but the same as yesterday applies Haven't checked it and don't even know if I understood the question:p: (Guess you'll have to do the question David if I've failed)

First, draw a nice little graph of two straight lines (one with negative and one with positive gradient) intersecting at p on the y-axis. The function is defined for values between 10 and -10 on the x-axis; The gradients will be p/10 and -p/10.

f(t)=\displaystyle\{^{-\frac{p}{10}t+p , -10&lt;t&lt;0}_{\frac{p}{10}t +p , 0&lt;t&lt;10}

\int_{-10}^0 -\frac{p}{10}(t+p) dt + \int_0^{10} \frac{p}{10}(t+p) dt =1 and by symmetry either of those parts equals 0.5.
Working with the left one we have -\frac{p}{10}(-50)=0.5 so p=1/10

1-P(miss the bus)=P(catch the bus)

P(miss)=(1/2)*P(bus arrives earlier than 7.55)+(1/2)(4/5)P(bus arrives earlier than 7.58)+(1/2)(1/5)*P(bus arrives earlier than 8.02)

P(bus arrives earlier than 7.55)=\frac{1}{100}\int_{-10}^{-5} (t+10) dt=\frac{25}{200}=\frac{1}{8}

P(bus arrives earlier than 7.58)=\frac{1}{100}\int_{-10}^{-2} (t+10)dt=\frac{32}{100}=\frac{8}  {25}

P(bus arrives earlier than 8.02)=0.5+(0.5-\frac{8}{25})=\frac{17}{25}

Multiplying this lot together \frac{1}{2}\frac{1}{8}+\frac{1}{  2}\frac{4}{5}\frac{8}{25}+\frac{  1}{2}\frac{1}{5}\frac{17}{25}=\f  rac{517}{2000}
So 1- 517/2000 = \frac{1483}{2000}


When buying a new clock he will arrive at 7.55 every day (according to that clock).
For him not to get on a bus by 9am it means he needs to miss the first bus (supposed to go at 8) and that the second bus (supposed to leave at 9) doesn't arrive in time. The probability the the second bus doesn't arrive in time is 1/2. Thus this probability is \frac{1}{2}\frac{1}{8}=\frac{1}{  16}=\frac{6}{96}

If it has happened that he misses the bus 5 times in 100 days, since \frac{5}{100}&lt;\frac{6}{96} it definitely suggests that the clock isn't behind. Possibly slightly ahead, although it is only deviating slightly from the expected number of missed buses.


edit: I think I've got II/10 (matrices) out, will maybe type it up tomorrow.
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Rabite
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I think I agree with whatever your solution was to III Q8 (The differential equation with Ps and Qs), but I didn't have time to finish it before my shift started. I got a 56e² though, which probably means we did the same thing.

Not sure if that has any bearing on the chance of it being the right solution though, hehe...
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Dystopia
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(Original post by Rabite)
I think I agree with whatever your solution was to III Q8 (The differential equation with Ps and Qs), but I didn't have time to finish it before my shift started. I got a 56e² though, which probably means we did the same thing.

Not sure if that has any bearing on the chance of it being the right solution though, hehe...
Hurray, my solution isn't necessarily dodgy!
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Dystopia
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STEP I, Q12

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Tension in spring using modulus of elasticity:
\displaystyle T_{1} = \frac{\lambda e}{l_{0}} = \frac{\lambda (2l - l\sqrt{3})}{2l} = \frac{\lambda (2 - \sqrt{3})}{2}

Tension in string by resolving vertically:

\displaystyle 2T_{2}\cos\frac{\pi}{3} = mg \Rightarrow T_{2} = mg

Tension in string by resolving horizontally:
\displaystyle T_{1} = T_{2}\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}T_{2}

\displaystyle \Rightarrow \frac{\lambda (2 - \sqrt{3})}{2} = \frac{\sqrt{3}}{2}mg

\displaystyle \Rightarrow mg = \lambda \left(\frac{2}{\sqrt{3}} - 1\right)

As required.

Using elastic potential energy = \displaystyle \frac{\lambda e^{2}}{2l_{0}}, we get \displaystyle EPE = \frac{\lambda (2l - 2xl)^{2}}{4l} = \lambda l (1 - x)^{2}

Using pythagoras, the change in height is -\sqrt{l^{2} - l^{2}x^{2}} = -l\sqrt{1 - x^{2}}

Change in gravitational potential energy = mgh = - mgl\sqrt{1-x^{2}}

Energy is conserved, so \lambda l (1-x)^{2} - mgl\sqrt{1-x^{2}} + \frac{1}{2}mv^{2} = 0

When the particle comes to rest, v = 0, so:

\displaystyle \lambda l (1-x)^{2} - mgl\sqrt{1-x^{2}} = 0

Substituting for mg, we get:
\displaystyle \Rightarrow \lambda (1-x)^{2} - \lambda \left(\frac{2}{\sqrt{3}} - 1\right)\sqrt{1-x^{2}} = 0

Note that in 1990 calculators were allowed, so it is easy to numerically confirm that a possible solution for x is approximately 0.66:

\displaystyle 0.34^{2} - \left(\frac{2}{\sqrt{3}} - 1\right)\sqrt{1 - 0.66^{2}} \approx -0.0006 \approx 0
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Dystopia
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STEP I, Q9

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\displaystyle y = \frac{1}{x}, \; \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{x^{2}}

Therefore the equation of the tangent at the point \displaystyle \left(b, \frac{1}{b}\right) is \displaystyle y = -\frac{1}{b^{2}}x + \frac{2}{b}

The co-ordinates of C are at the intersection of \displaystyle y = -x + 2 and \displaystyle y = -\frac{1}{b^{2}} + \frac{2}{b}

\displaystyle -x + 2 = -\frac{1}{b^{2}} + \frac{2}{b} \Rightarrow x\left(1-\frac{1}{b^{2}}\right) = 2\left(1-\frac{1}{b}\right)

\displaystyle x = \frac{2}{1 + \frac{1}{b}} = \frac{2b}{b+1}

\displaystyle y = \frac{2}{b+1}

Using the expression for the area of a trapezium:

Area of ACC'A' = \displaystyle \frac{1}{2}\left(1 + \frac{2}{b+1}\right)\left(\frac{  2b}{b+1}\right) = \frac{b(b+3)}{(b+1)^{2}}

Area of CBB'C' = \displaystyle \frac{1}{2}\left(\frac{2}{b+1} + \frac{1}{b}\right)\left(b - \frac{2b}{b+1}\right) = \frac{(3b+1)(b-1)}{2(b+1)^{2}}

Sum of Areas = \displaystyle \frac{2b(b+3) + (3b+1)(b-1)}{2(b+1)^{2}} = \frac{5b^{2} + 4b - 1}{2(b+1)^{2}} = \frac{5b - 1}{2(b+1)}

As the gradient is constantly increasing (becoming 'less' negative), the area under the curve between 1 and b is greater than the sum of the areas of ACC'A' and CBB'C:

\displaystyle \int^{b}_{1} \frac{1}{x} \; \mathrm{d}x = \ln b \geq \frac{5b-1}{2(b+1)}, \; b &gt; 1

\displaystyle \Rightarrow \ln (1 + z) \geq \frac{5z + 4}{2z}, \; z &gt; 0

To prove the left-hand inequality, it is sufficient to show that \displaystyle \frac{5z + 4}{2z} \geq \frac{2z}{2+z}, \; z &gt; 0

\displaystyle 5z^{2} + 14z + 8 \geq 4z^{2} \Leftrightarrow z^{2} + 14z + 8 \geq 0, \; z &gt; 0

Which is clearly true.

The inequality \displaystyle \ln(1+z) \leq z follows immediately from the fact that 1 + z \leq e^{z} by the Maclaurin Expansion of e^{z}.

Alternatively, note that the area of the trapezium ABB'A' is greater than the area under the curve between 1 and b.

\displaystyle \ln b \leq \frac{1}{2}\left(1 + \frac{1}{b}\right)\left(b-1 \right), \; b &gt; 1

\displaystyle \ln (1+z) \leq \frac{z^{2} + 2z}{2 + 2z}, \; z&gt;0

\displaystyle \frac{z^{2} + 2z}{2 + 2z} \leq z \Leftrightarrow z^{2} + 2z \leq 2z^{2} + 2z \Leftrightarrow 0 \leq z^{2}

Therefore \displaystyle \frac{2z}{2+z} \leq \ln (1+z) \leq z, \; z&gt;0
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STEP I, Q4

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Arithmetic Progressions
If, \displaystyle a_{1}, a_{2}, \; ... \;, a_{n} is an arithmetic progression, then

\displaystyle S_{n} = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(2a_{1} + (n-1)(a_{2}-a_{1})) = \frac{n}{2}((3-n)a_{1} + (n-1)a_{2})

Double Arithmetic Progressions
If \displaystyle b_{2}-b_{1}, \; b_{3}-b_{3}, \; ... \;, b_{n+1}-b_{n}, \; ... is an arithmetic progression, then

\displaystyle b_{n} = a + (n-2)d + b_{n-1}

\displaystyle \Rightarrow b_{n} = \left( \sum^{n}_{r=2} a + (r-2)d\right) + b_{1} = (n-1)a + d \left( \sum^{n}_{r=2}r-2 \right) + b_{1}

\displaystyle \sum^{n}_{r=2}(r-2) = \sum^{n-2}_{k=1} k = \frac{1}{2}(n-1)(n-2)

\displaystyle \Rightarrow b_{n} = (n-1)a + \frac{1}{2}(n-1)(n-2)d + b_{1}

\displaystyle \sum_{r=1}^{n} b_{r} = a\left(\sum_{r=1}^{n} n-1\right) + \frac{d}{2}\left(\sum_{r=1}^{n} n^{2} - 3n + 2 \right) + nb_{1}

\displaystyle \sum_{r=1}^{n} b_{r} = \frac{1}{2}n(n-1)a + \frac{1}{6}n(n-1)(n-2)d + nb_{1}

\displaystyle a = b_{2} - b_{1}, \; d = (b_{3} - b_{2}) - (b_{2} - b_{1}) = b_{3} - 2b_{2} + b_{1}

Several minutes of tedious algebra later:

\displaystyle \sum_{r=1}^{n} b_{r} = \frac{1}{6}n\left((n^{2} - 6n + 11)b_{1} - (n-1)(2n-7)b_{2} + (n-1)(n-2)b_{3}\right)

Factorial Progressions
b_{4}-b_{2} = d, \; b_{6}-b_{4} = 2d, \; 220 - b_{6} = 6d

b_{2} - 1 = a, \; b_{n} - b_{n-1} = a + (n-2)x

(b_{3} - b_{2}) + (b_{4} - b_{3}) = b_{4} - b_{2} = 2a + 3x = d

(b_{5}-b_{4}) + (b_{6} - b_{5}) = b_{6} - b_{4} = 2a + 7x = 2d

2a + 7x = 4a + 6x \Rightarrow x = 2a, \; d = 8a

(b_{2}-1) + (b_{3}-b_{2}) + \cdots + (b_{6}-b_{5}) = b_{6} - 1 = 5a + 10x

b_{6} = 220 - 6d = 5a + 10x + 1 \Rightarrow 5a + 10x + 6d = 219

5a + 20a + 48a = 219 \Rightarrow 73a = 219

\Rightarrow a = 3, \; x = 6, \; d = 24

Putting a = 3, d = 6, b1 = 1 into

\displaystyle \sum_{r=1}^{n} b_{r} = \frac{1}{2}n(n-1)a + \frac{1}{6}n(n-1)(n-2)d + nb_{1}

\displaystyle S_{n} = \frac{3}{2}n(n-1) + n(n-1)(n-2) + n = \frac{1}{2}n(2n^{2} - 3n + 3)
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brianeverit
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paper I/4 & 7
Comments and /or corrections welcomed.
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brianeverit
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(Original post by *bobo*)
STEP I
10) i)total resistive forces= 50+2(v-3)= 2(22 +v) where v is pos. in direction of turning point from starting point.
using P=Fv, where F=resistive forces due to no acceleration
300=2v(22+v)
2v^2 + 44v- 300=0
v=0.25 (in forward journey), v=1.25(backward journey) to 2dp
using v=s/t:
total time for race= 5000/0.25 + 5000/1.25
= 24000 seconds (supposed to answer to nearest second but don't have a calc on me)


ii) total energy expended using P=W/t
W=300x24000=7200000

x(1)t(1)+x(2)t(2)= 7200000

x(1)= v(1)(50+2(v(1)-3))= v(1)(44 + 2v)
t(1)= 5000/x(1)

x(2)= v(2)(50+2(v(2) +3))=v(2)(56+2v(2))
t(2)=5000/v(2)

=> 5000(44+2v(1))+5000(56+2v(2))=72 00000

=> 100+ 2v(1) + 2v(2)= 1440
=> v(1) +v(2)=670 so is independant of x(1) and x(2) values.
How did you egt v=0.25 for solution of v^2+11v-150=0?
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brianeverit
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Paper I/3
Copmments and /or corrections welcomed
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brianeverit
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Paper I/9
Confirmation of solution welcomed.
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brianeverit
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paper I/11,12 and 13
Confirmation of these solutions would be most welcome.
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Paper I number 15
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