Given that, for all \(x\), \[\frac{ax^2+bx+c}{(x-\alpha)(x-\beta)(x-\gamma)} = \frac{A}{x-\alpha} + \frac{B}{x-\beta}+ \frac{C}{x-\gamma},\] find the condition that \(A+B+C = 0\). Hence, or otherwise, evaluate \[\sum_{n=1}^{N} \frac{3n+1}{n(n+1)(n+2)}.\]
Solution: Proof 1: Let \(f(x) = \frac{ax^2+bx+c}{(x-\alpha)(x-\beta)(x-\gamma)}\) then \(\lim_{x \to \infty} xf(x) = a\) but also \(A+B+C\), therefore \(A+B+C = a\). Therefore \(A+B+C = 0 \Leftrightarrow a = 0\) Proof 2: Multiply both sides by \((x-\alpha)(x-\beta)(x-\gamma)\) then the coefficient of \(x^2\) is \(a\) on the LHS and \(A+B+C\) on the RHS, therefore \(A+B+C = a\) \begin{align*} && \frac{3n+1}{n(n+1)(n+2)} &= \frac{1}{2n} + \frac{2}{n+1} - \frac{5}{2(n+2)} \\ \Rightarrow && \sum_{n=1}^N \frac{3n+1}{n(n+1)(n+2)} &= \sum_{n=1}^N\left ( \frac{1}{2n} + \frac{2}{n+1} - \frac{5}{2(n+2)}\right) \\ &&&= \quad \frac{1}{2 \cdot 1} + \frac{2}{2} - \frac{5}{2 \cdot 3} + \cdots \\ &&&\quad\,\,\, + \frac{1}{2 \cdot 2} + \frac{2}{3} - \frac{5}{2 \cdot 4} + \cdots \\ &&&\quad\,\,\, + \frac{1}{2 \cdot 3} + \frac{2}{4} - \frac{5}{2 \cdot 5} + \cdots \\ &&&\quad \quad \quad \cdots \\ &&&\quad\,\,\, + \frac{1}{2 \cdot (N-2)} + \frac{2}{N-1} - \frac{5}{2 \cdot N} + \cdots \\ &&&\quad\,\,\, + \frac{1}{2 \cdot (N-1)} + \frac{2}{N} - \frac{5}{2 \cdot (N+1)} + \cdots \\ &&&\quad\,\,\, + \frac{1}{2 \cdot N} + \frac{2}{N+1} - \frac{5}{2 \cdot (N+2)} + \cdots \\ &&&= \frac3{2} +\frac{2}{N+1} - \frac{5}{2(N+1)} - \frac{5}{2(N+2)} \\ &&&= \frac3{2} - \frac{1}{2(N+1)} - \frac{5}{2(N+2)} \end{align*}
An equation has the property that if \(x\) is any (real or complex) root then \(1/x\) and \(1-x\) are also roots. What other expressions in \(x\) are also roots? A quintic equation without repeated roots has the above property. Determine the roots.
Show that there are less than 300 primes \(p\) with \(1000 \leq p \leq 2000\).
Let \(x\) be any real number. The symbol \([x]\) denotes the greatest integer less than or equal to \(x\) (e.g. \([-4\frac{1}{2}] = -5\), \([\pi] = 3\)). Which of the following statements are true and which false? Prove the true ones, and, for each of the false ones, give an example in which the statement fails to hold. (i) \([x+y] \geq [x]+[y]\). (ii) \([x^2] = [x]^2\). (iii) \([[x]/n] = [x/n]\) for every positive integer \(n\). (iv) \([nx] = n[x]\) for every positive integer \(n\). Show that \[\sum_{k=0}^{n-1} \left[ a+\frac{k}{n} \right] = [na]\] for any real number \(a\) and any integer \(n \geq 1\). [Hint: deal first with the case \(0 \leq a < 1/n\) and then consider the effect of replacing \(a\) by \(a+1/n\) and \(a-1/n\).]
The roots of the equation \(x^3 + ax^2 +bx+ c = 0\) are distinct and form a geometric progression. Taken in another order, they form an arithmetic progression. Find \(b\) and \(c\) in terms of \(a\).
Let \(\xi\) be any irrational number. Show that, given any integer \(a\), there is a unique integer \(b\) such that \(0 < a\xi+b < 1\). Use this to show that, given any positive integer \(N\), there are distinct numbers \(n\xi+m\) and \(n'\xi+m'\) (\(n\), \(m\), \(n'\), \(m'\) all integers) whose distance apart is less than \(1/N\). Hence deduce that there are integers \(r\) and \(s\) such that \(0 < r\xi+s < 1/N\). Suppose now that \(q\) is a rational number and \(N > 0\) an integer. Prove that there is a number of the form \(n\xi + m\) between \(q\) and \(q + 1/N\). [Hint: Suppose first that \(q > 0\), and that \(r\xi+s\) has been determined as above. Consider the first integer \(k\) such that \(kr\xi + ks > q\).]
Let \(G\) be a group. The centre of \(G\) is defined by \(Z = \{ x \in G: xg = gx \text{ for all } g \in G\}\). Show that \(Z\) is a subgroup of \(G\), and that \(Z\) is an abelian (i.e. commutative) group. Show that when \(G\) is the group of order 6 consisting of the rotations and reflections preserving an equilateral triangle, \(Z\) consists of the identity alone, but when \(G\) is the group of order 8 consisting of the rotations and reflections preserving a square, then \(Z\) has more than one element.
Let \(A\) be any \(2 \times 2\) matrix with integer entries. The trace of \(A\) is defined to be the sum of the diagonal elements of \(A\) \[e.g. \text{trace} \begin{pmatrix} 1 & 7\\ 4 & 6 \end{pmatrix} = 1 + 6 = 7.\] Show that the function \(f(A) = \text{trace } A\) satisfies the following rules: (i) \(f(A+B) = f(A)+f(B)\). (ii) \(f(\lambda A) = \lambda f(A)\) for any integer \(\lambda\). (iii) \(f(AB) = f(BA)\). (iv) If \(I\) is the \(2 \times 2\) identity matrix, then \(f(I) = 2\). Suppose \(f\) is any other function which also satisfies (i) to (iv). Let \(E_{ij}\) be the matrix with 1 in the \((ij)\) position and 0 elsewhere. Use rules (ii) and (iii) to prove that \[f(E_{12}) = f(E_{21}) = 0 \text{ and } f(E_{11}) = f(E_{22}).\] Hence use (i), (ii) and (iv) to deduce that \(f(A) = \text{trace } A\) for all \(2 \times 2\) integer matrices \(A\).
Show that \[\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}\] and hence, by induction or otherwise, evaluate \[\sum_{q=0}^{n} \binom{n+q}{q} \frac{1}{2^{n+q}}.\] [The binomial coefficient \(\binom{n}{r}\) is defined by \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\).]
Given a triangle \(ABC\) show that it is possible to construct three mutually touching circles with centres \(A\), \(B\), \(C\), respectively. What values are possible for the radius of the circle centre \(A\) in terms of the lengths \(a\), \(b\), \(c\) of the sides of the triangle?