Problems

2012 Paper 3 Q8

D: 1700.0 B: 1500.0

sequences recurrence relation Fibonacci sequence proof by induction arctan telescoping series trigonometric identity

The sequence $F_0$, $F_1$, $F_2$, $\ldots\,$ is defined by $F_0=0$, $F_1=1$ and, for $n\ge0$, \[ F_{n+2} = F_{n+1} + F_n \,. \]

Show that $F_0F_3-F_1F_2 = F_2F_5- F_3F_4\,$.
Find the values of $F_nF_{n+3} - F_{n+1}F_{n+2}$ in the two cases that arise.
Prove that, for $r=1$, $2$, $3$, $\ldots\,$, \[ \arctan \left( \frac 1{F_{2r}}\right) =\arctan \left( \frac 1{F_{2r+1}}\right)+ \arctan \left( \frac 1{F_{2r+2}}\right) \] and hence evaluate the following sum (which you may assume converges): \[ \sum_{r=1}^\infty \arctan \left( \frac 1{F_{2r+1}}\right) \,. \]

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2011 Paper 3 Q7

D: 1700.0 B: 1486.2

proof by induction surds binomial expansion integer expressions Pell equation algebraic identity nested radicals consecutive integers

Let \[ T _n = \left( \sqrt{a+1} + \sqrt a\right)^n\,, \] where $n$ is a positive integer and $a$ is any given positive integer.

In the case when $n$ is even, show by induction that $T_n$ can be written in the form \[ A_n +B_n \sqrt{a(a+1)}\,, \] where $A_n$ and $B_n$ are integers (depending on $a$ and $n$) and $A_n^2 =a(a+1)B_n^2 +1$.
In the case when $n$ is odd, show by considering $(\sqrt{a+1} +\sqrt a)T_m$ where $m$ is even, or otherwise, that $T_n$ can be written in the form \[ C_n \sqrt {a+1} + D_n \sqrt a \,, \] where $C_n$ and $D_n$ are integers (depending on $a$ and $n$) and $ (a+1)C_n^2 = a D_n^2 +1\,$.
Deduce that, for each $n$, $T_n$ can be written as the sum of the square roots of two consecutive integers.

Solution:

Claim: For all $n \geq 1$ $T_{2n} = A_{2n} + B_{2n}\sqrt{a(a+1)}$ where $A_{2n}, B_{2n}$ are integers and $A_{2n}^2 = a(a+1)B_{2n}^2+1$ Proof: (By induction) Base case: $n =1$. \begin{align*} && T_2 &= (\sqrt{a+1}+\sqrt{a})^2 \\ &&&= a+1+2\sqrt{a(a+1)}+a \\ &&&=2a+1+2\sqrt{a(a+1)} \\ \Rightarrow && A_2 &= 2a+1 \\ && B_2 &= 2 \\ && A_2^2 &= 4a^2+4a+1 \\ && a(a+1)B_1^2 + 1 &= 4a^2+4a+1 \end{align*} Therefore our base case is true. Suppose it is true for some $n = k$ then consider $n = k+1$ we must have $T_{2k} = A_{2k}+B_{2k}\sqrt{a(a+1)}$ \begin{align*} && T_{2(k+1)} &= T_{2k} (\sqrt{a+1}+\sqrt{a})^2 \\ &&&= \left (A_{2k}+B_{2k}\sqrt{a(a+1)} \right)\left (2a+1+2\sqrt{a(a+1)} \right) \\ &&&= (2a+1)A_{2k}+2a(a+1)B_{2k} + (2A_{2k}+(2a+1)B_{2k})\sqrt{a(a+1)} \\ \Rightarrow && A_{2(k+1)} &= (2a+1)A_{2k}+2a(a+1)B_{2k} \in \mathbb{Z} \\ && B_{2(k+1)} &= 2A_{2k}+(2a+1)B_{2k} \in \mathbb{Z} \\ && A_{2(k+1)}^2 &= \left ( (2a+1)A_{2k}+2a(a+1)B_{2k} \right)^2 \\ &&&= (2a+1)^2A_{2k}^2+4a^2(a+1)^2B_{2k}^2+4a(a+1)(2a+1)A_{2k}B_{2k} \\ &&&= a^2(a+1)^2(2a+1)^2B_{2k}^2+(2a+1)^2\\ &&&\quad\quad+4a^2(a+1)^2B_{2k}^2+4a(a+1)(2a+1)A_{2k}B_{2k}\\ && a(a+1)B_{2(k+1)}^2 + 1 &= a(a+1)\left ( 2A_{2k}+(2a+1)B_{2k} \right)^2 \\ &&&= 4a(a+1)A_{2k}^2 + 4a(a+1)(2a+1)A_{2k}B_{2k} + a(a+1)(2a+1)^2B_{2k}^2 + 1\\ &&&= 4a^2(a+1)^2B_{2k}^2+4a(a+1) + \\ &&&\quad\quad+ 4a(a+1)(2a+1)A_{2k}B_{2k} + a(a+1)(2a+1)^2B_{2k}^2 + 1 \end{align*} So our relation holds ad therefore by induction we're done.
When $n$ is odd, notice that \begin{align*} && T_n &= (\sqrt{a+1}+\sqrt{a})(A_m+B_m\sqrt{a(a+1)})\\ &&&= \sqrt{a+1}(\underbrace{A_m+aB_m}_{C_n})+\sqrt{a}(\underbrace{(a+1)B_m+A_m}_{D_n}) \\ \\ && (a+1)C_n^2 &= (a+1)(A_m+aB_m)^2 \\ &&&= (a+1)(A_m^2+2aA_mB_m+B_m^2) \\ &&&= (a+1)(a(a+1)B_m^2+1+2aA_mB_m+B_m^2) \\ &&&= a(a+1)^2B_m^2 + 2a(a+1)A_mB_m+(a+1)B_m^2+a+1 \\ && aD_n^2+1 &= a((a+1)^2B_m + 2(a+1)A_mB_m + A_m^2)+1 \\ &&&= a(a+1)^2B_m + 2a(a+1)A_mB_m + aB_{m}^2+a+1 \end{align*}
For even $n$ $T_n = \sqrt{a(a+1)B_{2n}^2+1} + \sqrt{a(a+1)B_{2n}^2}$ For odd $n$, $T_n = \sqrt{aD_n^2+1}+ \sqrt{aD_n^2}$ therefore it is always the sum of the square root of two consecutive integers.

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2010 Paper 3 Q7

D: 1700.0 B: 1516.0

differential equation Maclaurin series proof by induction Leibniz rule recurrence relation Chebyshev polynomial trigonometric higher derivatives

Given that $y = \cos(m \arcsin x)$, for $\vert x \vert <1$, prove that \[ (1-x^2) \frac {\d^2 y}{\d x^2} -x \frac {\d y}{\d x} +m^2y=0\,. \] Obtain a similar equation relating $\dfrac{\d^3y}{\d x^3}\,$, $\dfrac{\d^2y}{\d x^2}\, $ and $\, \dfrac{\d y}{\d x}\,$, and a similar equation relating $\dfrac{\d^4y}{\d x^4}\,$, $\dfrac{\d^3y}{\d x^3}\,$ and $\,\dfrac{\d^2 y}{\d x^2}\,$. Conjecture and prove a relation between $\dfrac{\d^{n+2}y}{\d x^{n+2}}\,$, $\dfrac{\d^{n+1}y}{\d x^{n+1}}\;$ and $\;\dfrac{\d^n y}{\d x^n}\,$. Obtain the first three non-zero terms of the Maclaurin series for $y$. Show that, if $m$ is an even integer, $\cos m\theta$ may be written as a polynomial in $\sin\theta$ beginning \[ 1 - \frac{m^2\sin^2\theta}{2!}+ \frac{m^2(m^2-2^2)\sin^4\theta}{4!} -\cdots \,. \, \tag{$\vert\theta\vert < \tfrac12 \pi$} \] State the degree of the polynomial.

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2009 Paper 3 Q7

D: 1700.0 B: 1485.5

second order differential equations recurrence relation proof by induction differentiation Leibniz rule polynomial rational function

The functions $\f_n(x)$ are defined for $n=0$, $1$, $2$, $\ldots$\, , by \[ \f_0(x) = \frac 1 {1+x^2}\, \qquad \text{and}\qquad \f_{n+1}(x) =\frac{\d \f_n(x)}{\d x}\,. \] Prove, for $n\ge1$, that \[ (1+x^2)\f_{n+1}(x) + 2(n+1)x\f_n(x) + n(n+1)\f_{n-1}(x)=0\,. \]
The functions $\P_n(x)$ are defined for $n=0$, $1$, $2$, $\ldots$\, , by \[ \P_n(x) = (1+x^2)^{n+1}\f_n(x)\,. \] Find expressions for $\P_0(x)$, $\P_1(x)$ and $\P_2(x)$. Prove, for $n\ge0$, that \[ \P_{n+1}(x) -(1+x^2)\frac {\d \P_n(x)}{\d x}+ 2(n+1)x \P_n(x)=0\,, \] and that $\P_n(x)$ is a polynomial of degree $n$.

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2008 Paper 3 Q2

D: 1700.0 B: 1555.2

sequences summation binomial theorem recurrence relation proof by induction polynomial power sums

Let $S_k(n) \equiv \sum\limits_{r=0}^n r^k\,$, where $k$ is a positive integer, so that \[ S_1(n) \equiv \tfrac12 n(n+1) \text{ and } S_2(n) \equiv \tfrac16 n(n+1)(2n+1)\,. \]

By considering $\sum\limits_{r=0}^n \left[ (r+1)^k-r^k\right]\, $, show that \[ kS_{k-1}(n)=(n+1)^k -(n+1) - \binom{k}{2} S_{k-2}(n) - \binom {k}{3} S_{k-3}(n) - \cdots - \binom{k}{k-1} S_{1}(n) \;. \tag{$*$} \] Obtain simplified expressions for $S_3(n)$ and $S_4(n)$.
Explain, using $(*)$, why $S_k(n)$ is a polynomial of degree $k+1$ in $n$. Show that in this polynomial the constant term is zero and the sum of the coefficients is 1.

Solution:

\begin{align*} &&(n+1)^k &= \sum_{r=0}^n \left [ (r+1)^k - r^k \right] \\ &&&= \sum_{r=0}^n \left [ \left ( \binom{k}{0}r^k+\binom{k}1r^{k-1} + \binom{k}{2}r^{k-2} + \cdots + \binom{k}{k} 1 \right) - r^k\right] \\ &&&= \sum_{r=0}^n \left ( \binom{k}1r^{k-1} + \binom{k}{2}r^{k-2} + \cdots + \binom{k}{k} 1 \right) \\ &&&=k \sum_{r=0}^n r^{k-1} + \binom{k}{2}\sum_{r=0}^nr^{k-2} + \cdots + \binom{k}{k} \sum_{r=0}^n 1 \\ &&&= kS_{k-1}(n) + \binom{k}2 S_{k-2}(n) + \cdots +\binom{k}{k-1}S_1(n) + (n+1) \\ \Rightarrow && k S_{k-1}(n) &= (n+1)^k -(n+1) -\binom{k}2 S_{k-2}(n) - \cdots -\binom{k}{k-1}S_1(n) \\ && 4S_3(n) &= (n+1)^4-(n+1) - \binom{4}{2} \frac{n(n+1)(2n+1)}{6} - \binom{4}{3} \frac{n(n+1)}{2} \\ &&&= (n+1) \left ( (n+1)^3-1 - n(2n+1)-2n \right) \\ &&&= (n+1) \left ( n^3+3n^2+3n+1-1 - 2n^2-3n \right) \\ &&&= (n+1) \left ( n^3+n^2 \right) \\ &&&= n^2(n+1)^2 \\ \Rightarrow && S_3(n) &= \frac{n^2(n+1)^2}{4} \\ \\ &&5S_4(n) &=(n+1)^5-(n+1) - \binom{5}{2} \frac{n^2(n+1)^2}4 - \binom{5}{3} \frac{n(n+1)(2n+1)}{6} - \binom{5}{4} \frac{n(n+1)}{2} \\ &&&= (n+1) \left ((n+1)^4 - 1-\frac{5n^2(n+1)}{2} - \frac{5n(2n+1)}{3} -\frac{5n}{2}\right)\\ &&&= \frac{n+1}{6} \left (6(n+1)^4-6-15n^2(n+1)-10n(2n+1)-15n \right) \\ &&&= \frac{n+1}{6} \left (6n^4+24n^3+36n^2+24n+6 -6-15n^3-15n^2-20n^2-10n-15n\right) \\ &&&= \frac{n+1}{6} \left (6n^4+9n^3+n^2-n\right) \\ &&&= \frac{(n+1)n(2n+1)(3n^2+3n-1)}{6} \\ \Rightarrow && S_4(n) &= \frac{(n+1)n(2n+1)(3n^2+3n-1)}{30} \end{align*}
Proceeding by induction, since $S_k(n)$ is a polynomial of degree $k+1$ for small $k$, we can see that \[ (k+1)S_k(n) = \underbrace{(n+1)^{k+1}}_{\text{poly deg }=k+1} - \underbrace{(n+1)}_{\text{poly deg}=1} - \underbrace{\binom{k+1}{2}S_{k-1}(n)}_{\text{poly deg}=k} - \underbrace{\cdots}_{\text{polys deg}< k} - \underbrace{\binom{k+1}{k} S_1(n)}_{\text{poly deg}=1}\] therefore $S_k(n)$ is a polynomial of degree $k+1$ (in fact with leading coefficient $\frac{1}{k+1}$. Since $S_k(0) = \sum_{r=0}^{0} r^k = 0$ there is no constant term, and since $S_k(1) = \sum_{r=0}^1 r^k = 1$ the sum of the coefficients is $1$

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2008 Paper 3 Q5

D: 1700.0 B: 1499.3

proof by induction recurrence relation sequences polynomial identity Chebyshev polynomials quadratic equation geometric progression

The functions ${\rm T}_n(x)$, for $n=0$, 1, 2, $\ldots\,$, satisfy the recurrence relation \[ {\rm T}_{n+1}(x) -2x {\rm T}_n(x) + {\rm T}_{n-1}(x) =0\, \ \ \ \ \ \ \ (n\ge1). \tag{$*$} \] Show by induction that \[ \left({\rm T}_n(x)\right)^2 - {\rm T}_{n-1}(x) {\rm T}_{n+1}(x) = \f(x)\,, \] where $\f(x) = \left({\rm T}_1(x)\right)^2 - {\rm T}_0(x){\rm T}_2(x)\,$. In the case $\f(x)\equiv 0$, determine (with proof) an expression for ${\rm T}_n(x)$ in terms of ${\rm T}_0(x)$ (assumed to be non-zero) and ${\rm r}(x)$, where ${\rm r}(x) = {\rm T}_1(x)/ {\rm T}_0(x)$. Find the two possible expressions for ${\rm r}(x)$ in terms of $x$. %Conjecture (without proof) the general form of the solution of $(*)$.

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2007 Paper 3 Q3

D: 1700.0 B: 1469.5

proof by induction Fibonacci sequence recurrence relation divisibility sequences algebraic identity

A sequence of numbers, $F_1, F_2, \ldots$, is defined by $F_1=1, F_2=1$, and \[ F_n=F_{n-1}+F_{n-2}\, \quad \text{for $n\ge 3$}. \]

Write down the values of $F_3, F_4, \ldots , F_8$.
Prove that $F^{\vphantom{2}}_{2k+3}F^{\vphantom{2}}_{2k+1} -F_{2k+2}^2 = -F^{\vphantom{2}}_{2k+2}F^{\vphantom{2}}_{2k}+F_{2k+1}^2\,$.
Prove by induction or otherwise that $F^{\vphantom{2}}_{2n+1}F^{\vphantom{2}}_{2n-1}-F^2_{2n}=1\,$ and deduce that $F^2_{2n}+1\,$ is divisible by $F^{\vphantom{2}}_{2n+1}\,.$
Prove that $F^2_{2n-1}+1\,$ is divisible by $F^{\vphantom{2}}_{2n+1}\,.$

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2007 Paper 3 Q5

D: 1700.0 B: 1516.0

hyperbolic functions differentiation proof by induction coth logarithmic differentiation higher derivatives pattern recognition

Let $y = \ln (x^2-1)\,$, where $x >1$, and let $r$ and $\theta$ be functions of $x$ determined by $r= \sqrt{x^2-1}$ and $\coth\theta= x$. Show that \[ \frac {\d y}{\d x} = \frac {2\cosh \theta}{r} \text{ and } \frac {\d^2 y}{\d x^2} = -\frac {2 \cosh 2\theta}{r^2}\,, \] and find an expression in terms of $r$ and $\theta$ for $\dfrac {\d^3 y}{\d x^3}\,$. Find, with proof, a similar formula for $\dfrac{\d^n y}{\d x^n}$ in terms of $r$ and $\theta$.

Solution: \begin{align*} && y &= \ln(x^2 -1) \\ && r &= \sqrt{x^2-1} \\ && \coth \theta &= x \\ && r &= \sqrt{\coth^2 \theta - 1} = \sqrt{\textrm{cosech}^2 \theta} = \textrm{cosech} \theta \\ && \frac{\d y}{\d x} &= \frac{2x}{x^2-1} \\ &&&= \frac{2 \coth \theta}{r^2} \\ &&&= \frac{2 \cosh \theta}{\sinh \theta \cdot r \cdot \textrm{cosech} \theta } \\ &&&= \frac{2 \cosh \theta}{r } \\ \\ && \frac{\d^2 y}{\d x^2} &= \frac{2(x^2-1)-4x^2}{(x^2-1)^2} \\ &&&= \frac{-2(1+x^2)}{r^2 \textrm{cosech}^2 r} \\ &&&= -\frac{2(1 + \coth^2 \theta) \sinh^2 \theta}{r^2} \\ &&&= -\frac{2(\sinh^2 \theta + \cosh^2 \theta)}{r^2} \\ &&&= -\frac{2 \cosh 2 \theta}{r^2} \\ \\ && \frac{\d^3 y}{\d x^3} &= \frac{-4x(x^2-1)^2-(-2x^2-2)\cdot2(x^2-1)\cdot 2x}{(x^2-1)^4} \\ &&&= \frac{-4x(x^2-1)+8x(x^2+1)}{(x^2-1)^3}\\ &&&= \frac{4x^3+12x}{(x^2-1)^3} \\ &&&=\frac{\sinh^3 \theta (4\coth^3 \theta + 12\coth \theta )}{r^3} \\ &&&=\frac{4\cosh^3 \theta + 12\cosh \theta \sinh^2 \theta}{r^3} \\ &&&= \frac{4 \cosh 3 \theta}{r^3} \\ \end{align*} Claim: $\frac{\d^n y}{\d x^n} = (-1)^{n+1}\frac{2(n-1)!\cosh n \theta}{r^n}$ Proof: By induction. Base cases already proven \begin{align*} \frac{\d r}{\d x} &= \frac{x}{\sqrt{x^2-1}} = \frac{\coth \theta}{\textrm{cosech} \theta} = \cosh \theta \\ \frac{\d \theta}{\d x} &= - \sinh^2 \theta \\ \\ \frac{\d^{n+1} y}{\d x^{n+1}} &= (-1)^{n+1}(n-1)!\frac{\d}{\d x} \left ( \frac{2\cosh n \theta}{r^n}\right) \\ &= (-1)^{n+1}\frac{2 n \sinh n \theta \cdot r^n \cdot \frac{\d \theta}{\d x}- 2\cosh n \theta \cdot nr^{n-1} \frac{\d r}{\d x} }{r^{2n}} \\ &= (-1)^{n+2}\frac{2n( \cosh n \theta\cosh \theta + r\sinh n \theta \sinh^2 \theta) }{r^{n+1}} \\ &= (-1)^{n+2}n!\frac{2\cosh(n+1) \theta }{r^{n+1}} \\ \end{align*} We can think of this as $\ln(x^2-1) = \ln(x+1)+\ln(x-1)$ and also note $x \pm 1 = \coth \theta \pm 1 = \frac{\cosh \theta \pm \sinh \theta}{\sinh \theta} = \frac{e^{\pm \theta}}{\sinh \theta}$ \begin{align*} && \frac{\d^n}{\d x^n} \ln(x^2-1) &= (n-1)!(-1)^{n-1} \left ( \frac{1}{(x+1)^n} + \frac{1}{(x-1)^n} \right) \\ &&&= (-1)^{n-1}(n-1)! \left ( \frac{\sinh^n \theta}{e^{n\theta}} + \frac{\sinh^n \theta}{e^{-n\theta}} \right) \\ &&&= (-1)^{n-1} (n-1)!2\cosh n \theta \cdot \sinh^n \theta \\ &&&= (-1)^{n-1}(n-1)! \frac{2 \cosh n \theta }{r^n} \end{align*}

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2006 Paper 2 Q1

D: 1600.0 B: 1485.5

sequences recurrence relation convergence periodic sequence quadratic equation fixed points proof by induction inequality

The sequence of real numbers $u_1$, $u_2$, $u_3$, $\ldots$ is defined by \begin{equation*} u_1=2 \,, \qquad\text{and} \qquad u_{n+1} = k - \frac{36}{u_n} \quad \text{for } n\ge1, \tag{$*$} \end{equation*} where $k$ is a constant.

Determine the values of $k$ for which the sequence $(*)$ is: (a) constant; (b) periodic with period 2; (c) periodic with period 4.
In the case $k=37$, show that $u_n\ge 2$ for all $n$. Given that in this case the sequence $(*)$ converges to a limit $\ell$, find the value of $\ell$.

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2006 Paper 2 Q2

D: 1600.0 B: 1500.0

exponential series inequality proof by induction factorial curve sketching stationary points logarithmic function differentiation bounding estimates

Using the series \[ \e^x = 1 + x +\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}+\cdots\,, \] show that $\e>\frac83$. Show that $n!>2^n$ for $n\ge4$ and hence show that $\e<\frac {67}{24}$. Show that the curve with equation \[ y= 3\e^{2x} +14 \ln (\tfrac43-x)\,, \qquad {x<\tfrac43} \] has a minimum turning point between $x=\frac12$ and $x=1$ and give a sketch to show the shape of the curve.

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2006 Paper 3 Q8

D: 1700.0 B: 1500.0

proof proof by induction polynomials differentiation product rule linearity operator abstract algebra

$\triangle$ is an operation that takes polynomials in $x$ to polynomials in $x$; that is, given any polynomial $\h(x)$, there is a polynomial called $\triangle \h(x)$ which is obtained from $\h(x)$ using the rules that define $\triangle$. These rules are as follows:

$\triangle x = 1\,$;
$\triangle \big( \f(x)+\g(x)\big) = \triangle \f(x) + \triangle\g(x)\,$ for any polynomials $\f(x)$ and $\g(x)$;
$\triangle \big( \lambda \f(x)\big) =\lambda \triangle \f(x)$ for any constant $\lambda$ and any polynomial $\f(x)$;
$\triangle \big( \f(x)\g(x)\big) = \f(x) \triangle \g(x) +\g(x)\triangle \f(x)$ for any polynomials $\f(x)$ and $\g(x)$.

Using these rules show that, if $\f(x)$ is a polynomial of degree zero (that is, a constant), then $\triangle \f(x) =0$. Calculate $\triangle x^2$ and $\triangle x^3$. Prove that $\triangle \h(x) \equiv \dfrac{\d \h(x)}{\d x \ \ \ }$ for any polynomial $\h(x)$. You should make it clear whenever you use one of the above rules in your proof. $\vphantom{\int}$

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2005 Paper 3 Q4

D: 1700.0 B: 1457.9

sequences recurrence relation proof by induction geometric sequence periodic sequence ratio closed form expression

The sequence $u_n$ ($n= 1, 2, \ldots$) satisfies the recurrence relation \[ u_{n+2}= \frac{u_{n+1}}{u_n}(ku_n-u_{n+1}) \] where $k$ is a constant. If $u_1=a$ and $u_2=b\,$, where $a$ and $b$ are non-zero and $b \ne ka\,$, prove by induction that \[ u_{2n}=\Big(\frac b a \Big) u_{2n-1} \] \[ u_{2n+1}= c u_{2n} \] for $n \ge 1$, where $c$ is a constant to be found in terms of $k$, $a$ and $b$. Hence express $u_{2n}$ and $u_{2n-1}$ in terms of $a$, $b$, $c$ and $n$. Find conditions on $a$, $b$ and $k$ in the three cases:

the sequence $u_n$ is geometric;
$u_n$ has period 2;
the sequence $u_n$ has period 4.

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2004 Paper 1 Q8

D: 1500.0 B: 1547.8

sequences recurrence relation convergence proof by induction inequality strictly increasing quadratic recurrence fixed points

A sequence $t_0$, $t_1$, $t_2$, $...$ is said to be strictly increasing if $t_{n+1} > t_n$ for all $n\ge{0}\,$.

The terms of the sequence $x_0\,$, $x_1\,$, $x_2\,$, $\ldots$ satisfy $$ \ds x_{n+1}=\frac{x_n^2 +6}{5} $$ for $n\ge{0}\,$. Prove that if $x_0 > 3$ then the sequence is strictly increasing.
The terms of the sequence $y_0\,$, $y_1\,$, $y_2\,$, $\ldots$ satisfy $$ \ds y_{n+1}= 5-\frac 6 {y_n} $$ for $n\ge{0}\,$. Prove that if $2 < y_0 < 3$ then the sequence is strictly increasing but that $y_n<3$ for all $n\,$.

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2003 Paper 1 Q1

D: 1484.0 B: 1484.0

proof by induction summation simultaneous equations polynomial sum of squares sum of cubes undetermined coefficients

It is given that $\sum\limits_{r=-1}^ {n} r^2$ can be written in the form $pn^3 +qn^2+rn+s\,$, where $p\,$, $q\,$, $r\,$ and $s$ are numbers. By setting $n=-1$, $0$, $1$ and $2$, obtain four equations that must be satisfied by $p\,$, $q\,$, $r\,$ and $s$ and hence show that \[ { \sum\limits_{r=0} ^n} r^2= {\textstyle \frac16} n(n+1)(2n+1)\;. \] Given that $\sum\limits_{r=-2}^ nr^3$ can be written in the form $an^4 +bn^3+cn^2+dn +e\,$, show similarly that \[ { \sum\limits_{r=0} ^n} r^3= {\textstyle \frac14} n^2(n+1)^2\;. \]

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2003 Paper 2 Q3

D: 1600.0 B: 1541.7

proof by contradiction proof by induction irrational numbers cube root sequences convergence integer limit

Prove that the cube root of any irrational number is an irrational number. Let $\displaystyle u_n = {5\vphantom{\dot A}}^{1/{(3^n)}}\,$. Given that $\sqrt[3]5$ is an irrational number, prove by induction that $u_n$ is an irrational number for every positive integer $n$. Hence, or otherwise, give an example of an infinite sequence of irrational numbers which converges to a given integer $m\,$. [An irrational number is a number that cannot be expressed as the ratio of two integers.]

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