Question from the 2011 IMC (International Mathematics Competition) Key Stage III paper, about the evaluation of a quadratic equation

contest-mathproblem solvingquadraticssequences-and-seriestelescopic-series

When $a=1, 2, 3, …, 2010, 2011$, the roots of the equation $x^2-2x-a^2-a=0$ are $(a_1, b_1), (a_2, b_2), (a_3, b_3),\cdots, (a_{2010}, b_{2010}), (a_{2011}, b_{2011})$ respectively. Evaluate:

$$
\frac{1}{a_1} + \frac{1}{b_1} + \frac{1}{a_1} + \frac{1}{b_2} + \frac{1}{a_3} + \frac{1}{b_3} +\cdots + \frac{1}{a_{2010}} + \frac{1}{b_{2010}} + \frac{1}{a_{2011}} + \frac{1}{b_{2011}}
$$

I tried solving this question, with the use of the quadratic equation.

Using the quadratic equation, I concluded that $a_1=\frac{2+\sqrt{12}}{2}=1+\sqrt{3}$ and that $b_1=\frac{2-\sqrt{12}}{2}=1-\sqrt{3}$. The reason that I concluded to this is because the quadratic equation states:

$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

So I substituted $a$ with $1$ ($1$ is multiplying $x^2$, in the original equation), $b$ with $-2$ ($x$ is getting multiplied by $-2$ in the original equation) and $c$ with $(-a^2-a)$, as they are the only ones which are not getting directly multiplied by $x$ in the original equation.

Hence, I subsequently worked out that $\frac{1}{a_1} + \frac{1}{b_1} = \frac{2}{-2}=-1$

Continuing to do the same thing I worked out $\frac{1}{a_2} + \frac{1}{b_2}= \frac{2}{-6} = -\frac{1}{3}$ and $\frac{1}{a3} + \frac{1}{b_3}= \frac{2}{-12}=-\frac{1}{6}$ and $\frac{1}{a_4} + \frac{1}{b_4} = \frac{2}{-20}=-\frac{1}{10}$ and $\frac{1}{a_5} + \frac{1}{b5} = \frac{2}{-30} =-\frac{1}{15}$.

I subsequently realised that a pattern was emerging, the denominator, each time is getting increased by the degree of $n$ at which $a$ and $b$ are (for instance $\frac{1}{a_4} + \frac{1}{b_4}= -\frac{1}{6+4}$)

Had I not been dealing with fractions, I would have solved it using arithmetic progressions, but unfortunately that is not possible.

I can think of no other way of finishing off my thoughts, nor any other way to solve this problem. Can you please help me? Can you please tell me if there is any method of finishing off my thoughts and if there isn't, can you please suggest a method of solving the problem

Thank you in advance

Best Answer

$$\sum_{k=1}^{2011}\left(\frac{1}{a_k}+\frac{1}{b_k}\right)=\sum_{k=1}^{2011}\frac{a_k+b_k}{a_kb_k}=\sum_{k=1}^{2011}\frac{2}{-k(k+1)}=$$ $$=-2\sum_{k=1}^{2011}\left(\frac{1}{k}-\frac{1}{k+1}\right)=-2\left(1-\frac{1}{2012}\right)=-\frac{2011}{1006}.$$

Related Solutions

[Math] The quadratic spline

Here is a cubic-spline interpolation for the $4$ points given in your question:

$ f(x)= \begin{cases} +0.1 \overline3(x-3.0)^3 -1.3 (x-3.0)+2.5 & \text{$3.0 \leq x \leq 4.5$}\\ -0.15\overline1(x-4.5)^3+0.6 (x-4.5)^2+0.0\overline4(x-4.5)+1.0 & \text{$4.5 \leq x \leq 7.0$}\\ +0.0 \overline8(x-7.0)^3-0.5\overline3(x-7.0)^2-0.2\overline8(x-7.0)+2.5 & \text{$7.0 \leq x \leq 9.0$}\\ \end{cases} $

Here is a piece of C code for any given number of points $(x_0,y_0),(x_1,y_1),\ldots,(x_N,y_N)$:

void Spline(double x[N+1],double y[N+1], // input
            double A[N],double B[N],     // output
            double C[N],double D[N])     // output
{
    double w[N];
    double h[N];
    double ftt[N+1];

    for (int i=0; i<N; i++)
    {
        w[i] = (x[i+1]-x[i]);
        h[i] = (y[i+1]-y[i])/w[i];
    }

    ftt[0] = 0;
    for (int i=0; i<N-1; i++)
        ftt[i+1] = 3*(h[i+1]-h[i])/(w[i+1]+w[i]);
    ftt[N] = 0;

    for (int i=0; i<N; i++)
    {
        A[i] = (ftt[i+1]-ftt[i])/(6*w[i]);
        B[i] = ftt[i]/2;
        C[i] = h[i]-w[i]*(ftt[i+1]+2*ftt[i])/6;
        D[i] = y[i];
    }
}

void PrintSpline(double x[N+1],           // input
                 double A[N],double B[N], // input
                 double C[N],double D[N]) // input
{
    for (int i=0; i<N; i++)
    {
        printf("%f <= x <= %f : f(x) = ",x[i],x[i+1]);
        printf("%f(x-%f)^3 + ",A[i],x[i]);
        printf("%f(x-%f)^2 + ",B[i],x[i]);
        printf("%f(x-%f) + "  ,C[i],x[i]);
        printf("%f\n"         ,D[i]);
    }
}

Here is a piece of Python code for any given number of points $(x_0,y_0),(x_1,y_1),\ldots,(x_N,y_N)$:

class Point:
    def __init__(self,x,y):
        self.x = 1.0*x
        self.y = 1.0*y

def Spline(points):
    N   = len(points)-1
    w   =     [(points[i+1].x-points[i].x)      for i in range(0,N)]
    h   =     [(points[i+1].y-points[i].y)/w[i] for i in range(0,N)]
    ftt = [0]+[3*(h[i+1]-h[i])/(w[i+1]+w[i])    for i in range(0,N-1)]+[0]
    A   =     [(ftt[i+1]-ftt[i])/(6*w[i])       for i in range(0,N)]
    B   =     [ftt[i]/2                         for i in range(0,N)]
    C   =     [h[i]-w[i]*(ftt[i+1]+2*ftt[i])/6  for i in range(0,N)]
    D   =     [points[i].y                      for i in range(0,N)]
    return A,B,C,D

def PrintSpline(points,A,B,C,D):
    for i in range(0,len(points)-1):
        func = str(points[i].x)+' <= x <= '+str(points[i+1].x)+' : f(x) = '
        components = []
        if A[i]:
            components.append(str(A[i])+'(x-'+str(points[i].x)+')^3')
        if B[i]:
            components.append(str(B[i])+'(x-'+str(points[i].x)+')^2')
        if C[i]:
            components.append(str(C[i])+'(x-'+str(points[i].x)+')')
        if D[i]:
            components.append(str(D[i]))
        if components:
            func += components[0]
            for i in range (1,len(components)):
                if components[i][0] == '-':
                    func += ' - '+components[i][1:]
                else:
                    func += ' + '+components[i]
            print func
        else:
            print func+'0'

def Example():
    points = [Point(3.0,2.5),Point(4.5,1.0),Point(7.0,2.5),Point(9.0,0.5)]
    A,B,C,D = Spline(points)
    PrintSpline(points,A,B,C,D)

Please note that the two pieces of code above assume $x_0 < x_1 < \ldots < x_N$.

Question from the 2011 IMC (international mathematics competition) key stage III paper, about a logical sequence

In a group of $a$ liars and $b$ truth-tellers, each truth-teller will identify $a$ liars and each liar will "identify" $b$ liars. Hence in total, there will be $2ab$ answers "liar". Today, we have $$ 2ab=240$$ and yesterday, we had $$2(a-1)b=216\qquad\text{or}\qquad 2a(b-1)=216.$$ By subtracting we get $2b=240-216=24$ (and thereby $a=10$, $a+b=22$) or $2a=24$ (and thereby $b=10$ and again $a+b=22$).

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