Throughout this question, \(N\) is an integer with \(N \geqslant 1\) and \(S_N = \displaystyle\sum_{r=1}^{N} \frac{1}{r^2}\).
You may assume that \(\displaystyle\lim_{N\to\infty} S_N\) exists and is equal to \(\frac{1}{6}\pi^2\).
- Show that
\[\frac{1}{r+1} - \frac{1}{r} + \frac{1}{r^2} = \frac{1}{r^2(r+1)}.\]
Hence show that
\[\sum_{r=1}^{N} \frac{1}{r^2(r+1)} = \sum_{r=1}^{N} \frac{1}{r^2} - 1 + \frac{1}{N+1}.\]
Show further that \(\displaystyle\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} = \frac{1}{6}\pi^2 - 1\).
- Find \(\displaystyle\sum_{r=1}^{N} \frac{1}{r^2(r+1)(r+2)}\) in terms of \(S_N\), and hence evaluate \(\displaystyle\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)(r+2)}\).
- Show that
\[\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)^2} = \sum_{r=1}^{\infty} \frac{2}{r^2(r+1)} - 1.\]
Show Solution
- \(\,\) \begin{align*}
&& \frac1{r+1} - \frac1r + \frac1{r^2} &= \frac{-1}{r(r+1)} + \frac{1}{r^2} \\
&&&= \frac{r+1-r}{r^2(r+1)} \\
&&&= \frac{1}{r^2(r+1)}
\end{align*}
Therefore
\begin{align*}
&& \sum_{r=1}^N \frac1{r^2(r+1)} &= \sum_{r=1}^N \left (\frac1{r+1} - \frac1r + \frac1{r^2} \right) \\
&&&= \sum_{r=1}^N \left (\frac1{r+1} - \frac1r\right) + \sum_{r=1}^N \frac1{r^2} \\
&&&=\frac{1}{N+1} - 1 + \sum_{r=1}^N \frac1{r^2} \\
\end{align*}
therefore
\begin{align*}
&& \sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} &= \lim_{N \to \infty } \sum_{r=1}^{N} \frac{1}{r^2(r+1)} \\
&&&= \lim_{N \to \infty } \left (\frac{1}{N+1} - 1 + \sum_{r=1}^N \frac1{r^2} \right) \\
&&&= -1 +\lim_{N \to \infty } \sum_{r=1}^N \frac1{r^2} \\
&&&= -1 + \sum_{r=1}^\infty \frac1{r^2} \\
&&&= \frac{\pi^2}{6}-1
\end{align*}
- Note that \begin{align*}
&& \frac{1}{r^2(r+1)(r+2)} &= \frac{Ar+B}{r^2} + \frac{C}{r+1} + \frac{D}{r+2} \\
&&&= \frac{1}{2r^2} + \frac{1}{r+1} - \frac{1}{4(r+2)} - \frac{3}{4r}
\end{align*}
So
\begin{align*}
&& \sum_{r=1}^N \frac{1}{r^2(r+1)(r+2)} &= \sum_{r=1}^N \left ( \frac{1}{2r^2} + \frac{1}{r+1} - \frac{1}{4(r+2)} - \frac{3}{4r} \right ) \\
&&&= \frac12 \sum_{r=1}^N \frac{1}{r^2} + \frac{1}{2} - \frac14 \cdot \frac1{3} - \frac34 \frac11 + \\
&&& \quad \quad \quad + \frac13 - \frac14\frac14 - \frac34\frac12 + \\
&&& \quad \quad \quad + \frac14 - \frac14\frac15 - \frac34\frac13 + \\
&&&\quad \quad \quad+ \cdots + \\
&&&\quad \quad \quad + \frac1{N+1} - \frac14\frac1{N+2} - \frac34\frac1N \\
&&&= \frac12 \sum_{r=1}^N \frac{1}{r^2} +\frac14\frac12 - \frac34\frac11-\frac14\frac1{N+2}+\frac34 \frac1{N+1} \\
\\
\Rightarrow && \sum_{r=1}^{\infty} \frac{1}{r^2(r+1)(r+2)} &= \lim_{N \to \infty} \left [ \frac12 \sum_{r=1}^N \frac{1}{r^2} +\frac14\frac12 - \frac34\frac11-\frac14\frac1{N+2}+\frac34 \frac1{N+1}\right] \\
&&&= \frac{\pi^2}{12} -\frac58
\end{align*}
- Notice that \(\frac{1}{r^2(r+1)^2} - \frac{2}{r^2(r+1)} = \frac{1-2(r+1)}{r^2(r+1)^2} = \frac{-1-2r}{r^2(r+1)^2} = \frac{1}{(r+1)^2} - \frac{1}{r^2}\) and so
\begin{align*}
&& \sum_{r=1}^N \frac{1}{r^2(r+1)^2} &= \sum_{r=1}^N \left ( \frac{2}{r^2(r+1)} +\frac{1}{(r+1)^2} - \frac{1}{r^2} \right) \\
&&&= \sum_{r=1}^N \frac{2}{r^2(r+1)} +\frac{1}{(N+1)^2} - 1 \\
\end{align*}
and the result follows as \(N \to \infty\)
[There is a beautiful paper by KConrad about this question: https://kconrad.math.uconn.edu/blurbs/analysis/series_acceleration.pdf]
In this question, you need not consider issues of convergence.
For positive integer \(n\) let
\[\mathrm{f}(n) = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \ldots\]
and
\[\mathrm{g}(n) = \frac{1}{n+1} - \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} - \ldots\,.\]
- Show, by considering a geometric series, that \(0 < \mathrm{f}(n) < \dfrac{1}{n}\).
- Show, by comparing consecutive terms, that \(0 < \mathrm{g}(n) < \dfrac{1}{n+1}\).
- Show, for positive integer \(n\), that \((2n)!\,\mathrm{e} - \mathrm{f}(2n)\) and \(\dfrac{(2n)!}{\mathrm{e}} + \mathrm{g}(2n)\) are both integers.
- Show that if \(q\,\mathrm{e} = \dfrac{p}{\mathrm{e}}\) for some positive integers \(p\) and \(q\), then \(q\,\mathrm{f}(2n) + p\,\mathrm{g}(2n)\) is an integer for all positive integers \(n\).
- Hence show that the number \(\mathrm{e}^2\) is irrational.
A sequence \(u_n\), where \(n = 1, 2, \ldots\), is said to have \emph{degree} \(d\) if \(u_n\), as a function of \(n\), is a polynomial of degree \(d\).
- Show that, in any sequence \(u_n\) \((n = 1, 2, \ldots)\) that satisfies \(u_{n+1} = \frac{1}{2}(u_{n+2} + u_n)\) for all \(n \geqslant 1\), there is a constant difference between successive terms.
Deduce that any sequence \(u_n\) for which \(u_{n+1} = \frac{1}{2}(u_{n+2} + u_n)\), for all \(n \geqslant 1\), has degree at most 1.
- The sequence \(v_n\) \((n = 1, 2, \ldots)\) satisfies \(v_{n+1} = \frac{1}{2}(v_{n+2} + v_n) - p\) for all \(n \geqslant 1\), where \(p\) is a non-zero constant. By writing \(v_n = t_n + pn^2\), show that the sequence \(v_n\) has degree 2.
Given that \(v_1 = v_2 = 0\), find \(v_n\) in terms of \(n\) and \(p\).
- The sequence \(w_n\) \((n = 1, 2, \ldots)\) satisfies \(w_{n+1} = \frac{1}{2}(w_{n+2} + w_n) - an - b\) for all \(n \geqslant 1\), where \(a\) and \(b\) are constants with \(a \neq 0\). Show that the sequence \(w_n\) has degree 3.
Given that \(w_1 = w_2 = 0\), find \(w_n\) in terms of \(n\), \(a\) and \(b\).
The Fibonacci numbers are defined by \(F_0 = 0\), \(F_1 = 1\) and, for \(n \geqslant 0\), \(F_{n+2} = F_{n+1} + F_n\).
- Prove that \(F_r \leqslant 2^{r-n} F_n\) for all \(n \geqslant 1\) and all \(r \geqslant n\).
- Let \(S_n = \displaystyle\sum_{r=1}^{n} \frac{F_r}{10^r}\).
Show that
\[\sum_{r=1}^{n} \frac{F_{r+1}}{10^{r-1}} - \sum_{r=1}^{n} \frac{F_r}{10^{r-1}} - \sum_{r=1}^{n} \frac{F_{r-1}}{10^{r-1}} = 89S_n - 10F_1 - F_0 + \frac{F_n}{10^n} + \frac{F_{n+1}}{10^{n-1}}\,.\]
- Show that \(\displaystyle\sum_{r=1}^{\infty} \frac{F_r}{10^r} = \frac{10}{89}\) and that \(\displaystyle\sum_{r=7}^{\infty} \frac{F_r}{10^r} < 2 \times 10^{-6}\). Hence find, with justification, the first six digits after the decimal point in the decimal expansion of \(\dfrac{1}{89}\).
- Find, with justification, a number of the form \(\dfrac{r}{s}\) with \(r\) and \(s\) both positive integers less than \(10000\) whose decimal expansion starts
\[0.0001010203050813213455\ldots\,.\]
A sequence \(u_1, u_2, \ldots, u_n\) of positive real numbers is said to be unimodal if there is a value \(k\) such that
\[u_1 \leqslant u_2 \leqslant \ldots \leqslant u_k\]
and
\[u_k \geqslant u_{k+1} \geqslant \ldots \geqslant u_n.\]
So the sequences \(1, 2, 3, 2, 1\);\ \(1, 2, 3, 4, 5\);\ \(1, 1, 3, 3, 2\) and \(2, 2, 2, 2, 2\) are all unimodal, but \(1, 2, 1, 3, 1\) is not.
A sequence \(u_1, u_2, \ldots, u_n\) of positive real numbers is said to have property \(L\) if \(u_{r-1}u_{r+1} \leqslant u_r^2\) for all \(r\) with \(2 \leqslant r \leqslant n-1\).
- Show that, in any sequence of positive real numbers with property \(L\),
\[u_{r-1} \geqslant u_r \implies u_r \geqslant u_{r+1}.\]
Prove that any sequence of positive real numbers with property \(L\) is unimodal.
- A sequence \(u_1, u_2, \ldots, u_n\) of real numbers satisfies \(u_r = 2\alpha u_{r-1} - \alpha^2 u_{r-2}\) for \(3 \leqslant r \leqslant n\), where \(\alpha\) is a positive real constant. Prove that, for \(2 \leqslant r \leqslant n\),
\[u_r - \alpha u_{r-1} = \alpha^{r-2}(u_2 - \alpha u_1)\]
and, for \(2 \leqslant r \leqslant n-1\),
\[u_r^2 - u_{r-1}u_{r+1} = (u_r - \alpha u_{r-1})^2.\]
Hence show that the sequence consists of positive terms and is unimodal, provided \(u_2 > \alpha u_1 > 0\).
In the case \(u_1 = 1\) and \(u_2 = 2\), prove by induction that \(u_r = (2-r)\alpha^{r-1} + 2(r-1)\alpha^{r-2}\).
Let \(\alpha = 1 - \dfrac{1}{N}\), where \(N\) is an integer with \(2 \leqslant N \leqslant n\).
In the case \(u_1 = 1\) and \(u_2 = 2\), prove that \(u_r\) is largest when \(r = N\).
A sequence \(u_k\), for integer \(k \geqslant 1\), is defined as follows.
\[ u_1 = 1 \]
\[ u_{2k} = u_k \text{ for } k \geqslant 1 \]
\[ u_{2k+1} = u_k + u_{k+1} \text{ for } k \geqslant 1 \]
- Show that, for every pair of consecutive terms of this sequence, except the first pair, the term with odd subscript is larger than the term with even subscript.
- Suppose that two consecutive terms in this sequence have a common factor greater than one. Show that there are then two consecutive terms earlier in the sequence which have the same common factor. Deduce that any two consecutive terms in this sequence are co-prime (do not have a common factor greater than one).
- Prove that it is not possible for two positive integers to appear consecutively in the same order in two different places in the sequence.
- Suppose that \(a\) and \(b\) are two co-prime positive integers which do not occur consecutively in the sequence with \(b\) following \(a\). If \(a > b\), show that \(a-b\) and \(b\) are two co-prime positive integers which do not occur consecutively in the sequence with \(b\) following \(a-b\), and whose sum is smaller than \(a+b\). Find a similar result for \(a < b\).
- For each integer \(n \geqslant 1\), define the function \(\mathrm{f}\) from the positive integers to the positive rational numbers by \(\mathrm{f}(n) = \dfrac{u_n}{u_{n+1}}\). Show that the range of \(\mathrm{f}\) is all the positive rational numbers, and that \(\mathrm{f}\) has an inverse.
- The coefficients in the series
\[
S= \tfrac13 x + \tfrac 16 x^2 + \tfrac1{12} x^3 + \cdots + a_rx^r +
\cdots
\]
satisfy a recurrence relation of the form \(a_{r+1} + p a_r =0\). Write
down the value of \(p\).
By considering \((1+px)S\), find an expression for the sum to infinity
of \(S\) (assuming that it exists). Find also
an expression for the sum of the first \(n+1\) terms of \(S\).
- The coefficients in the series
\[
T=2 + 8x +18x^2+37 x^3 +\cdots + a_rx^r + \cdots
\]
satisfy a recurrence relation of the form \(a_{r+2}+pa_{r+1} +qa_r=0\).
Find an expression for the sum to infinity of \(T\) (assuming that it
exists). By expressing \(T\) in partial fractions, or
otherwise,
find
an expression for the sum of the first \(n+1\) terms of \(T\).
The sequence \(a_n\) is defined by \(a_0 = 1\) , \(a_1 = 1\) , and
$$
a_n = {1 + a_{n - 1}^2 \over a_{n - 2} } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( n \ge 2 ) .
$$
Prove by induction that
$$
a_n = 3 a_{n - 1} - a_{n - 2} \ \ \ \ \ \ \ \ \ \ \ ( n \ge2 ) .
$$
Hence show that
$$
a_n = {\alpha^{2 n - 1} + \alpha^{- ( 2 n - 1 )} \over \sqrt 5}
\ \ \ \ \ \ (n\ge1),
$$
where \(\displaystyle{\alpha = {1 + \sqrt 5 \over 2}}\).
The value \(V_N\) of a bond after \(N\) days is determined by the equation
$$
V_{N+1} = (1+c) V_{N} -d \qquad (c>0, \ d>0),
$$
where \(c\) and \(d\) are given constants.
By looking for solutions of the form \(V_T= A k^T + B\) for some constants \(A,B\) and \(k\), or otherwise, find \(V_N\) in terms of \(V_0\).
What is the solution for \(c=0\)? Show that this is the limit (for fixed \(N\)) as \(c\rightarrow 0\) of your solution for \(c>0\).
Show Solution
Suppose \(V_T = Ak^T + B\), then
\begin{align*}
&& Ak^{T+1}+B &= (1+c)(Ak^T+B) - d \\
\Rightarrow && (k-1-c)Ak^T &= cB -d \\
\Rightarrow && k &= 1+c \\
&& B &= \frac{d}{c} \\
&& A &= V_0 - B \\
\Rightarrow && V_N &= (V_0 - \frac{d}{c})(1+c)^{N} + \frac{d}{c}
\end{align*}
When \(c = 0\), \(V_{N+1} = V_N - d \Rightarrow V_N = V_0 - Nd\).
\begin{align*}
\lim_{c \to 0} \left ( (V_0 - \frac{d}{c})(1+c)^{N} + \frac{d}{c} \right) &= \lim_{c \to 0} \left ( \frac{(V_0 c-d)(1+c)^N + d}{c} \right ) \\
&= \lim_{c \to 0} \left ( \frac{V_0c - d-Ncd+NV_0c^2 + o(c^2) + d}{c} \right ) \\
&= \lim_{c \to 0} \left ( V_0 - Nd + o(c) \right ) \\
&= V_0 - Nd
\end{align*}
The real numbers \(u_{0},u_{1},u_{2},\ldots\) satisfy the difference
equation
\[
au_{n+2}+bu_{n+1}+cu_{n}=0\qquad(n=0,1,2,\ldots),
\]
where \(a,b\) and \(c\) are real numbers such that the quadratic equation
\[
ax^{2}+bx+c=0
\]
has two distinct real roots \(\alpha\) and \(\beta.\) Show that the above difference equation is satisfied by the numbers \(u_{n}\) defined
by
\[
u_{n}=A\alpha^{n}+B\beta^{n},
\]
where
\[
A=\frac{u_{1}-\beta u_{0}}{\alpha-\beta}\qquad\mbox{ and }\qquad B=\frac{u_{1}-\alpha u_{0}}{\beta-\alpha}.
\]
Show also, by induction, that these numbers provide the only solution.
Find the numbers \(v_{n}\) \((n=0,1,2,\ldots)\) which satisfy
\[
8(n+2)(n+1)v_{n+2}-2(n+3)(n+1)v_{n+1}-(n+3)(n+2)v_{n}=0
\]
with \(v_{0}=0\) and \(v_{1}=1.\)
Show Solution
First notice that \(u_n = \alpha^n\) and \(u_n = \beta^n\) both satisfy the recurrence, since:
\begin{align*}
&& a \alpha^2 + b \alpha + c &= 0 \\
\Rightarrow && a \alpha^{n+2} + b \alpha^{n+1} + c \alpha^n &= 0 \\
\Rightarrow && a u_{n+2} + bu_{n+1} + cu_n &=0
\end{align*}
Notice also that if \(u_n\) and \(v_n\) both satisfy the recurrence, then any linear combination of them will satisfy the recurrence:
\begin{align*}
&& \begin{cases} au_{n+2} + bu_{n+1} + cu_n &= 0 \\
av_{n+2} + bv_{n+1} + cv_n &= 0 \\ \end{cases} \\
\Rightarrow && a (\lambda u_{n+2}+ \mu v_{n+2}) + b (\lambda u_{n+1}+ \mu v_{n+1}) + c (\lambda u_{n}+ \mu v_{n}) &= 0
\end{align*}
by adding a linear combination of the top two equations.
Therefore it suffices to check that the constants \(A\) and \(B\) are such that we match \(u_0\) and \(u_1\).
\(\frac{u_1 - \beta u_0}{\alpha - \beta} + \frac{u_1 - \alpha u_0}{\beta - \alpha} = u_0\) and \(\frac{u_1 - \beta u_0}{\alpha - \beta}\alpha + \frac{u_1 - \alpha u_0}{\beta - \alpha}\beta = u_1\). So we are done.
Suppose we have another sequence, then we first notice that the first and second terms must be identical to each other. Suppose the first \(k\) terms are identical, then since the \(k+1\)th term depends only on the \(k\) and \(k-1\)th terms (both of which are equal) the \(k+1\)th term is the same. Therefore, by the principle of mathematical induction, all terms are the same.
First notice that if you put \(v_n = (n+1)w_n\) we have
\begin{align*}
&& 8(n+3)(n+2)(n+1)w_{n+2} - 2(n+3)(n+2)(n+1)w_{n+1} - (n+3)(n+2)(n+1)w_n &= 0 \\
\Rightarrow && 8w_{n+2}-2w_{n+1}-w_n &= 0
\end{align*}
This has characteristic equation \(8\lambda^2 - 2\lambda - 1 = 0 \Rightarrow \lambda = \frac12, -\frac14\).
Therefore the general solution is \(w_n = A \l \frac12 \r^n + B \l -\frac14\r^n\) and \(v_n = (n+1)\l A \l \frac12 \r^n + B \l -\frac14\r^n \r\).
When \(n = 0\) we have \(A+B = 0 \Rightarrow B =-A\).
When \(n=1\) we have \(1 = 2 \l \frac{A}{2} + \frac{A}{4} \r \Rightarrow A = \frac{4}{3}\), therefore
\[ v_n = \frac{4}{3}(n+1) \l \frac{1}{2^n} + \l -\frac14\r^n \r\]
Each day, books returned to a library are placed on a shelf in order of arrival, and left there. When a book arrives for which there is no room on the shelf, that book and all books subsequently returned are put on a trolley. At the end of each day, the shelf and trolley
are cleared. There are just two-sizes of book: thick, requiring two units of shelf space; and thin, requiring one unit. The probability that a returned book is thick is \(p\), and the probability that it is thin is \(q=1-p.\) Let \(M(n)\) be the expected number of books that
will be put on the shelf, when the length of the shelf is \(n\) units and \(n\) is an integer, on the assumption that more books will be returned each day than can be placed on the shelf. Show, giving reasoning,
that
- \(M(0)=0;\)
- \(M(1)=q;\)
- \(M(n)-qM(n-1)-pM(n-2)=1,\) for \(n\geqslant2.\)
Verify that a possible solution to these equations is
\[
M(n)=A(-p)^{n}+B+Cn,
\]
where \(A,B\) and \(C\) are numbers independent of \(n\) which you should express in terms of \(p\).
Show Solution
- \(M(0) = 0\) since if there's no space on the shelf, we wont be able to put any books on the shelf.
- If the shelf has length \(1\) it can only fit a thin book. For a thin book to be placed on the shelf, the very first book which comes to be placed must be thin. But this happens with probability \(q\). Therefore \(M(1) = q\).
- Suppose no books have been placed on the shelf, then with probability \(p\) a large book gets placed on the shelf, and the expected number of books to be placed on the shelf is equivalent to how many books will be placed on the shelf if the shelf only had \(n-2\) spaces. This is \(M(n-2)\). Similar if the book which arrives first is thin (with probability \(q\)) then there will be \(M(n-1)\) more books placed on the shelf in expectation. We've just added \(1\) more book, therefore \(M(n) = 1+pM(n-2) + qM(n-1)\) or rearranging \(M(n) - qM(n-1) - pM(n-2) = 1\).
Suppose \(M(n) = (-p)^n\), notice that:
\begin{align*}
M(n) - qM(n-1) - pM(n-2) &= (-p)^n - (1-p)(-p)^n - p(-p)^{n-2} \\
&= (-p)^{n-2}(p^2+(1-p)p-p) \\
&= 0
\end{align*}
Suppose \(M(n) = B\), notice that:
\begin{align*}
M(n) - qM(n-1) - pM(n-2) &= B - (1-p)B - pB \\
&= 0
\end{align*}
Finally, if \(M(n) = Cn\) notice that:
\begin{align*}
M(n) - qM(n-1) - pM(n-2) &= Cn - (1-p)C(n-1) - pC(n-2) \\
&= C(n(1-(1-p)+p)+(1-p)+2p) \\
&= C(1+p)
\end{align*}
Therefore if \(C = \frac{1}{1+p}\) we have that:
\(M(n) = A(-p)^n + B + Cn\) satisfies our recurrence.
We also need \(M(0) = 0\) and \(M(1) = q\)
\begin{align*}
0 &= M(0) \\
&= A + B \\
1-p &= M(1) \\
&= -pA+B
\end{align*}
\((1+p)A = p-1 \Rightarrow A = \frac{p-1}{1+p}, B = \frac{1-p}{1+p}\).
Therefore:
\[ M(n) = -\frac{1-p}{1+p}(-p)^n + \frac{1-p}{1+p} + \frac{n}{1+p} \]
is a possible solution to this equation
Let \(a\) and \(b\) be positive integers such that \(b<2a-1\). For any given positive integer \(n\), the integers \(N\) and \(M\) are defined by
\[
[a+\sqrt{a^{2}-b}]^{n}=N-r,
\]
\[
[a-\sqrt{a^{2}-b}]^{n}=M+s,
\]
where \(0\leqslant r<1\) and \(0\leqslant s<1\). Prove that
\begin{questionparts}
\item \(M=0\),
\item \(r=s\),
\item \(r^{2}-Nr+b^{n}=0.\)
\end{questionpart}
Show that for large \(n\), \(\left(8+3\sqrt{7}\right)^{n}\) differs from an integer by about \(2^{-4n}\).
Show Solution
- If we can show that \(0 < a - \sqrt{a^2-b} < 1\) then we will be done, since raising a number in \([0,1)\) to a positive integer power will always remain in the same interval.
Clearly \(\sqrt{a^2-b} < \sqrt{a^2} = a\) so we have \(a-\sqrt{a^2-b} > 0\)
We also have that \(\sqrt{a^2-b} > \sqrt{a^2-(2a-1)} = (a-1)\). Therefore \(a - \sqrt{a^2-b} < a - (a-1) = 1\) as required.
- If we can show that \(\l a + \sqrt{a^2-b} \r^n + \l a - \sqrt{a^2-b} \r^n = N -r + s\) is an integer we will be done, since the only integer value \(-r+s\) can be is \(0\).
This is easy to see, since
\begin{align*}
\l a + \sqrt{a^2-b} \r^n + \l a - \sqrt{a^2-b} \r^n &= \sum_{k=0}^n \binom{n}{k}a^{n-k}(\sqrt{a^2-b})^k +\sum_{k=0}^n \binom{n}{k}a^{n-k}(-\sqrt{a^2-b})^k \\
&= \sum_{k=0}^n \binom{n}{k}a^{n-k}\l (\sqrt{a^2-b})^k +(-\sqrt{a^2-b})^k \r \\
\end{align*}
But every term where \(k\) is odd in this sum is \(0\) (since they cancel) and ever term where \(k\) is even in this sum is an integer. Therefore the sum is an integer and we're done.
- \begin{align*}
-r^2+rN &= -r(r-N) \\
&= s(r-N) \\
&=- \l a - \sqrt{a^2-b} \r^n \l a + \sqrt{a^2-b} \r^n \\
&= -\l \l a - \sqrt{a^2-b} \r \l a + \sqrt{a^2-b} \r\r^n \\
&= - \l a^2 - a^2+b\r^n \\
&= b^n
\end{align*}
Therefore \(r^2-rN + b^n = 0\)
Looking at \(\left(8+3\sqrt{7}\right)^{n}\) we have \(a = 8, b = 1\) (since \(8^2 - 1 = 9 \cdot 7\). So we can apply the result of the previous question to see that:
\(\left(8+3\sqrt{7}\right)^{n}\) differs from an integer by \(\left(8-3\sqrt{7}\right)^{n}\).
\begin{align*}
8-3\sqrt{7} &= \frac{1}{8+3\sqrt{7}} \\
&\approx \frac{1}{8 + 8} \\
&\approx 2^{-4}
\end{align*}
Therefore it differs by approximation \((2^{-4})^n = 2^{-4n}\)