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1. By considering the binomial expansion of \((1+x)^{2m+1}\), prove that \[ \binom{ 2m \! +\! 1}{ m} < 2^{2m}\,, \] for any positive integer \(m\).
2. For any positive integers \(r\) and \(s\) with \(r< s\), \(P_{r,s}\) is defined as follows: \(P_{r,s}\) is the product of all the prime numbers greater than \(r\) and less than or equal to \(s\), if there are any such primes numbers; if there are no such primes numbers, then \(P_{r,s}=1\,\). For example, \(P_{3,7}=35\), \(P_{7,10}=1\) and \(P_{14,18}=17\). Show that, for any positive integer \(m\), \(P_{m+1\,,\, 2m+1} \) divides \(\displaystyle \binom{ 2m \! +\! 1}{ m} \,,\) and deduce that \[ P_{m+1\,,\, 2m+1} < 2^{2m} \,. \]
3. Show that, if \(P_{1,k} < 4^k\) for \(k = 2\), \(3\), \(\ldots\), \(2m\), then \( P_{1,2m+1} < 4^{2m+1}\,\).
4. Prove that \(\P_{1,n} < 4^n\) for \(n\ge2\).

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Solution:

1. Notice that \((1+x)^{2m+1} = 1+\binom{2m+1}{1}x+\cdots + \binom{2m+1}{m}x^{m} + \binom{2m+1}{m+1} + \cdots\). Notice also that \(\binom{2m+1}{m} = \binom{2m+1}{m+1}\). Therefore evaluating at \(x = 1\), we see \(2^{2m+1} > \binom{2m+1}{m} + \binom{2m+1}{m+1} = 2 \binom{2m+1}{m} \Rightarrow \binom{2m+1}{m} < 2^{2m}\)
2. Each prime dividing \(P_{m+1, 2m+1}\) divides the numerator of \(\binom{2m+1}{m}\) since it appears in \((2m+1)!\), but not the denominator, since they wont appear in \(m!\) or \((m+1)!\), and since they are prime they have to appear to divide it. Therefore the must divide \(\binom{2m+1}{m}\) and therefore \(P_{m+1,2m+1}\) must divide that binomail coefficient. Since \(a \mid b \Rightarrow a \leq b\) we must have \(P_{m+1, 2m+1} \leq \binom{2m+1}{m} < 2^{2m}\)
3. Since \begin{align*} P_{1,2m+1} &= P_{1,m+1}P_{m+1, 2m+1} \tag{split into primes below \(m+1\) and abvoe} \\ &< 4^{m+1}P_{m+1,2m+1} \tag{use the condition from the question}\\ &<4^{m+1}2^{2m} \tag{use our inequality} \\ &= 4^{2m+1} \end{align*}
4. We proceed by (strong) induction. Base case: (\(n = 2\)): \(P_{1,2} = 2 < 4^2 =16\) Suppose it is true for all \(k=2,3,\cdots,2m\) then it is true for \(k=2m+1\) by the previous part of the question. However it is also true for \(k=2m+2\), since that can never be prime (as n is now an even number bigger than 2). Therefore by the principle of mathematical induction it is true for all \(n\).

1. Show that if the acute angle between straight lines with gradients \(m_1\) and \(m_2\) is \(45^\circ\), then \[\frac{m_1 - m_2}{1 + m_1 m_2} = \pm 1.\]

The curve \(C\) has equation \(4ay = x^2\) (where \(a \neq 0\)).

2. If \(p \neq q\), show that the tangents to the curve \(C\) at the points with \(x\)-coordinates \(p\) and \(q\) meet at a point with \(x\)-coordinate \(\frac{1}{2}(p+q)\). Find the \(y\)-coordinate of this point in terms of \(p\) and \(q\). Show further that any two tangents to the curve \(C\) which are at \(45^\circ\) to each other meet on the curve \((y+3a)^2 = 8a^2 + x^2\).
3. Show that the acute angle between any two tangents to the curve \(C\) which meet on the curve \((y+7a)^2 = 48a^2 + 3x^2\) is constant. Find this acute angle.