The two sequences \(a_0\), \(a_1\), \(a_2\), \(\ldots\)
and
\(b_0\), \(b_1\), \(b_2\), \(\ldots\)
have general terms
\[
a_n = \lambda^n +\mu^n
\text { \ \ \ and \ \ \ }
b_n = \lambda^n - \mu^n\,,
\]
respectively, where \(\lambda = 1+\sqrt2\) and \(\mu= 1-\sqrt2\,\).
- Show that $\displaystyle \sum_{r=0}^nb_r = -\sqrt2 + \frac
1 {\sqrt2} \,a_{\low n+1}\,$,
and give a corresponding result for
\(\displaystyle \sum_{r=0}^na_r\,\).
- Show that, if \(n\) is odd,
$$\sum_{m=0}^{2n}\left( \sum_{r=0}^m a_{\low r}\right)
= \tfrac12 b_{n+1}^2\,,$$
and give a corresponding result when \(n\)
is even.
- Show that, if \(n\) is even,
$$\left(\sum_{r=0}^na_r\right)^{\!2}
-\sum_{r=0}^n a_{\low 2r+1} =2\,,$$
and give a corresponding result when
\(n\) is odd.
The first four terms of a sequence
are given by \(F_0=0\), \(F_1=1\), \(F_2=1\) and \(F_3=2\). The general term
is given by
\[
F_n= a\lambda^n+b\mu^n\,,
\tag{\(*\)}
\]
where \(a\), \(b\), \(\lambda\) and \(\mu\) are independent of \(n\), and \(a\) is
positive.
- Show that
\(\lambda^2 +\lambda\mu+ \mu^2 = 2\), and find the values of
\(\lambda\), \(\mu\), \(a\) and \(b\).
- Use \((*)\) to evaluate \(F_6\).
- Evaluate
\(\displaystyle \sum_{n=0}^\infty \frac{F_n}{2^{n+1}}\,.\)
Show Solution
- \(\,\)
\begin{align*}
&& 0 &= a+b \tag{1}\\
&& 1 &= a\lambda -a\mu \tag{2} \\
&& 1 &= a\lambda^2 -a\mu^2 \tag{3} \\
&& 2 &= a\lambda^3 - a\mu^3 \tag{4} \\
(4) \div (2): && 2 & = \lambda^2+\lambda \mu + \mu^2 \\
(3) \div (2): && 1 &= \lambda + \mu \\
\Rightarrow && 2 &= \lambda^2 + \lambda(1-\lambda) + (1-\lambda)^2 \\
&&&= \lambda^2-\lambda+1\\
\Rightarrow && \lambda, \mu &= \frac{1 \pm \sqrt{5}}{2} \\
\Rightarrow && a &= \frac{1}{\lambda - \mu} = \frac{1}{\sqrt{5}} \\
\Rightarrow && b &= -\frac{1}{\sqrt{5}}
\end{align*}
(NB: This is Binet's formula)
- \(\,\) \begin{align*} F_6 &= \frac{1}{\sqrt{5}} \left ( \left ( \frac{1 +\sqrt{5}}{2} \right)^6- \left ( \frac{1 -\sqrt{5}}{2} \right)^6 \right) \\
&= \frac{1}{2^6 \sqrt{5}} \left ( (1+\sqrt{5})^6-(1-\sqrt{5})^6 \right) \\
&= \frac{1}{2^5 \cdot \sqrt{5}} \left (6 \sqrt{5} +\binom{6}{3} (\sqrt{5})^3+\binom{6}{5}(\sqrt{5})^5 \right)\\
&= \frac{1}{2^5} \left (6 +20\cdot 5+6\cdot 5^2 \right)\\
&= \frac{1}{2^5} 256 = 2^3 = 8
\end{align*}
(way more painful than just computing it by adding terms!)
- \(\,\)
\begin{align*}
&& \sum_{n=0}^{\infty} \frac{F_n}{2^{n+1}} &= \sum_{n=0}^{\infty} \frac{a\lambda^n + b\mu^n}{2^{n+1}} \\
&&&= \frac12 \left ( \frac{a}{1-\frac{\lambda}2} + \frac{b}{1-\frac{\mu}2} \right) \\
&&&= \frac12 \left ( \frac{2a}{2-\lambda} + \frac{2b}{2-\mu}\right) \\
&&&= \frac{2a}{4-2\lambda} + \frac{2b}{4-2\mu}\\
&&&= \frac{2a}{4-(1+\sqrt{5})} - \frac{2a}{4-(1-\sqrt{5})} \\
&&&= \frac{2}{3\sqrt{5}-5} - \frac{2}{3\sqrt{5}+5} \\
&&&= \frac{6\sqrt{5}+10-6\sqrt{5}+10}{45-25} \\
&&&= 1
\end{align*}
Let \(x_{\low1}\), \(x_{\low2}\), \ldots, \(x_n\) and
\(x_{\vphantom {\dot A} n+1}\) be any fixed real numbers.
The numbers \(A\) and \(B\) are defined by
\[
A = \frac 1 n
\sum_{k=1}^n x_{ \low k}
\,, \ \ \
B= \frac 1 n
\sum_{k=1}^n (x_{\low k}-A)^2
\,, \ \ \
\]
and the numbers \(C\) and \(D\) are defined by
\[
C = \frac 1 {n+1}
\sum\limits_{k=1}^{n+1} x_{\low k}
\,,
\ \ \
D = \frac1{n+1}
\sum_{k=1}^{n+1} (x_{\low k}-C)^2
\,.
\]
- Express \( C\) in terms of \(A\), \(x_{\low n+1}\) and \(n\).
- Show that $ \displaystyle
B= \frac 1 n
\sum_{k=1}^n x_{\low k}^2
- A^2\,\(.
- Express \)D \( in terms of \)B\(, \)A\(, \)x_{\low n+1}\( and \)n$.
Hence show that \((n + 1)D \ge nB\)
for all values of \(x_{\low n+1}\), but that \(D < B\)
if and only if
\[
A-\sqrt{\frac{(n+1)B}{n}} < x_{\low n+1} <
A+\sqrt{\frac{(n+1)B}{n}}\,.
\]
The Fibonacci sequence \(F_1\), \(F_2\), \(F_3\), \(\ldots\) is defined by \(F_1=1\), \(F_2= 1\) and
\[
F_{n+1} = F_n+F_{n-1} \qquad\qquad (n\ge 2).
\]
Write down the values of \(F_3\), \(F_4\), \(\ldots\), \(F_{10}\). Let \(\displaystyle S=\sum_{i=1}^\infty \dfrac1 {F_i}\,\).
- Show that \(\displaystyle \frac 1{F_i} > \frac1{2F_{i-1}}\,\) for \(i\ge4\) and deduce that \(S > 3\,\). Show also that \(S < 3\frac23\,\).
- Show further that \(3.2 < S < 3.5\,\).
Show Solution
\begin{array}{c|r}
n & F_n \\ \hline
1 & 1 \\
2 & 1 \\
3 & 2 \\
4 & 3 \\
5 & 5 \\
6 & 8 \\
7 & 13 \\
8 & 21 \\
9 & 34 \\
10 & 55
\end{array}
\begin{questionparts}
\item Claim: \(\frac1{F_i} > \frac1{2F_{i-1}}\) for \(i \geq 4\).
Proof: Since \(F_i = F_{i-1}+F_{i-2}\) and \(F_i > 1\) for \(i \geq 1\) we have \(F_i > F_{i-1}\) for \(i \geq 3\). In particular we have \(F_i = F_{i-1}+F_{i-2} < 2F_{i-1}\) for \(i -1 \geq 3\) or \(i \geq 4\).
Therefore \(\frac{1}{F_i} > \frac1{2F_{i-1}}\)
I borrow \(C\) pounds at interest rate \(100\alpha \,\%\) per year.
The interest is added at the end of each year. Immediately after the
interest is added, I make
a repayment. The amount I repay at the end of the \(k\)th year is \(R_k\)
pounds and the amount I owe
at the beginning of \(k\)th year is \(C_k\) pounds (with \(C_1=C\)).
Express \(C_{n+1}\) in terms of \(R_k\) (\(k= 1\), \(2\), \(\ldots\), \(n\)), \(\alpha\) and \(C\)
and show that, if I pay off the loan in \(N\) years with repayments
given by \(R_k= (1+\alpha)^kr\,\), where \(r\) is constant, then \(r=C/N\,\).
If instead I pay off the loan in \(N\) years with \(N\) equal repayments
of \(R\) pounds, show that
\[
\frac R C = \frac{\alpha (1+\alpha)^{N} }{(1+\alpha)^N-1} \;,
\]
and that \(R/C\approx 27/103\) in the case \(\alpha =1/50\), \(N=4\,\).
How many integers greater than or equal to zero and less than a million are not divisible by 2 or 5? What is the average value of these integers?
How many integers greater than or equal to zero and less than 4179 are not divisible by 3 or 7? What is the average value of these integers?
Show Solution
There are \(1\,000\,000\) numbers between 1 and a million (inclusive). \(500\,000\) are divisible by \(2\), \(200\,000\) are divisible by \(5\) and \(100\,000\) are divisible by both. Therefore there are:
\(1\,000\,000 - 500\,000-200\,000+100\,000 = 400\,000\).
(Alternatively, the only numbers are those which are \(1,3,7,9 \pmod{10}\) so there are \(4\) every \(10\), or \(4 \cdot 100\,000\)).
We can sum all these values similarly,
\begin{align*}
S &= \underbrace{\sum_{i=1}^{10^6} i}_{\text{all numbers}}-\underbrace{\sum_{i=1}^{5 \cdot 10^5} 2i}_{\text{all multiples of } 2}-\underbrace{\sum_{i=1}^{2 \cdot 10^5} 5i}_{\text{all multiples of } 5}+\underbrace{\sum_{i=1}^{10^5} 10i}_{\text{all multiples of } 5} \\
&= \frac{10^6 \cdot (10^6 + 1)}{2} - \frac{10^6 \cdot (5\cdot 10^5+1)}{2} - \frac{10^6 \cdot (2\cdot 10^5+1)}{2} + \frac{10^6 \cdot (10^5+1)}{2} \\
&= \frac{10^6 (10^5 \cdot (10-5-2+1))))}{2} \\
&= \frac{10^6 \cdot 10^5 \cdot 4}{2} \\
&= 2\cdot 10^{11}
\end{align*}
So the average value is \(\frac{2 \cdot 10^{11}}{4 \cdot 10^5} = \frac{10^6}{2} = 500\,000\).
(Alternatively, each value can be paired off eg \(999\,999\) with \(1\) and so on, leaving averages of \(500\,000\)).
Note that \(4197\) is divisible by \(3\) and \(7\).
Using the same long we have:
\(4179 - \frac{4179}{3} - \frac{4179}{7} + \frac{4179}{21} = 4179 - 1393 - 597 + 199 = 2388\).
The sum will be:
\begin{align*}
S &= \underbrace{\sum_{i=1}^{4179}i }_{\text{all numbers}}- \underbrace{\sum_{i=1}^{1393}3i }_{\text{multiples of }3}- \underbrace{\sum_{i=1}^{597}7i }_{\text{multiples of }7}+ \underbrace{\sum_{i=1}^{199}21i }_{\text{mulitples of }21} \\
&= \frac{4179 \cdot 4180}{2} - \frac{4179 \cdot 1394}{2} - \frac{4179 \cdot 598}{2} +\frac{4179 \cdot 200}{2} \\
&= \frac{4179 \cdot 2388}{2}
\end{align*}
So the average value is \(\frac{4179}{2}\).
My bank pays \(\rho\%\) interest at the end of each year. I start with nothing in my account. Then for \(m\) years I deposit \(\pounds a\) in my account at the beginning of each year. After the end of the \(m\)th year, I neither deposit nor withdraw for \(l\) years.
Show that the total amount in my account at the end of this period is \[\pounds a\frac{r^{l+1}(r^{m}-1)}{r-1}\] where \(r=1+{\displaystyle \frac{\rho}{100}}\).
At the beginning of each of the \(n\) years following this period I withdraw \(\pounds b\) and this leaves my account empty after the \(n\)th withdrawal.
Find an expression for \(a/b\) in terms of \(r\), \(l\), \(m\) and \(n\).
Show Solution
Rather than putting the deposits in the same account, imagine they are all put in separate accounts. Then for example, the first \(\pounds a\) will go on to become \(\pounds r^m \cdot r^l a\) from \(m\) years of compound interest as more money is deposited, followed by \(l\) years where no money is deposited.
Therefore the total amount at the end is:
\begin{align*}
&& S &= r^{m}r^l a + r^{m-1}r^l a + \cdots + r r^l a \\
&&&= r^l a(r^m + \cdots + r) \\
&&&= ar^l\frac{r^{m+1}-r}{r-1} \\
&&&= a \frac{r^{l+1}(r^m-1)}{r-1}
\end{align*}
Rather than withdrawing \(b\) each time, imagine that in the \(n\)th year we withdraw each \(b\) with the appropriate additional interest, ie
\begin{align*}
&& \underbrace{a \frac{r^{l+1}(r^m-1)}{r-1}}_{\text{amount before \(n\) years}} \underbrace{r^{n-1}}_{\text{accounting for interest}} &= b r^{n-1} + br^{n-2} + \cdots + b \\
&&&= b \frac{r^n-1}{r-1} \\
\Rightarrow && \frac{a}{b} &= \frac{r^n-1}{r^{l+n}(r^m-1)}
\end{align*}
Find the sum of those numbers between 1000 and 6000 every one of whose digits
is one of the numbers \(0,\,2,\,5\) or \(7\), giving
your answer as a product of primes.
Show Solution
The first digit is \(2\) or \(5\), all the other digits can be any value from \(0,2,5,7\).
Therefore we have
\begin{align*}
S &= 2000 \cdot 4^3+5000 \cdot 4^3 + (200+500+700) \cdot 2 \cdot 4^2 + (20+50+70) \cdot 2 \cdot 4^2 + (2+5+7) \cdot 2 \cdot 4^2 \\
&= 7 \cdot 4^3 \cdot 2^3 \cdot 5^3 + 14 \cdot 2 \cdot 4^2 \cdot 111 \\
&= 2^{9} \cdot5^3 \cdot 7 + 2^{6} \cdot 3 \cdot 7 \cdot 37 \\
&= 2^6 \cdot 7 \cdot (1000+111) \\
&= 2^6 \cdot 7 \cdot 11 \cdot 101
\end{align*}
Alternatively, consider adding the first and last terms, and second and second and last terms, etc we obtain \(7777\). There are \(2 \cdot 4^3\) terms so \(7777 \cdot 4^3 = 2^6 \cdot 7 \cdot 11 \cdot 101\)
- By using the formula for the sum of a geometric series, or
otherwise, express the number \(0.38383838\ldots\) as a fraction in
its lowest terms.
- Let \(x\) be a real number which has a recurring decimal expansion
\[
x=0\cdot a_{1}a_{2}a_{2}\cdots,
\]
so that there exists positive integers \(N\) and \(k\) such that \(a_{n+k}=a_{n}\)
for all \(n>N.\) Show that
\[
x=\frac{b}{10^{N}}+\frac{c}{10^{N}(10^{k}-1)}\,,
\]
where \(b\) and \(c\) are integers to be found. Deduce that \(x\) is
rational.
Show Solution
- \(\,\) \begin{align*}
&& 0.383838\ldots &= \frac{3}{10} + \frac{8}{100} + \frac{3}{1000} + \cdots \\
&&&= \frac{38}{100} + \frac{38}{10000} + \cdots \\
&&&= \frac{38}{100} \left (1 + \frac{1}{100} + \frac{1}{100^2} + \cdots \right) \\
&&&= \frac{\frac{38}{100}}{1 - \frac{1}{100}} \\
&&&= \frac{38}{99}
\end{align*}
- Let \(x = 0\cdot a_{1}a_{2}a_{2}\cdots\) such that there exists \(N\), \(k\) such that \(a_{n+k} = a_n\) for \(n > N\).
First notice that
\begin{align*}
x &= \frac{a_1a_2\cdots a_N}{10\ldots000} + \frac{a_{N+1}}{10^{N+1}} + \cdots \\
&= \frac{b}{10^N} + \frac{a_{N+1}a_{N+2}\cdots a_{N+k}}{10^{N+k}} + \frac{a_{N+1}a_{N+2}\cdots a_{N+2k}}{10^{N+k}} + \cdots \\
&= \frac{b}{10^N} + \frac{c}{10^{N+k}} \left (1 + \frac{1}{10^k} + \cdots \right) \\
&= \frac{b}{10^N} + \frac{c}{10^{N+k}} \frac{1}{1- \frac{1}{10^k}} \\
&= \frac{b}{10^N} + \frac{c}{10^{N+k}} \frac{10^k}{10^k-1} \\
&= \frac{b}{10^N} + \frac{c}{10^N(10^k-1)}
\end{align*}
where \(b = a_1a_2\cdots a_N\) and \(c = a_{N+1}a_{N+2}\cdots a_{N+k}\). Clearly as the sum of two rational numbers, \(x\) is rational.
- If \(\mathrm{f}(r)\) is a function defined
for \(r=0,1,2,3,\ldots,\) show that
\[
\sum_{r=1}^{n}\left\{ \mathrm{f}(r)-\mathrm{f}(r-1)\right\} =\mathrm{f}(n)-\mathrm{f}(0).
\]
- If \(\mathrm{f}(r)=r^{2}(r+1)^{2},\) evaluate \(\mathrm{f}(r)-\mathrm{f}(r-1)\)
and hence determine \({\displaystyle \sum_{r=1}^{n}r^{3}.}\)
- Find the sum of the series \(1^{3}-2^{3}+3^{3}-4^{3}+\cdots+(2n+1)^{3}.\)
Show Solution
- \(\,\)
\begin{align*}
&& \sum_{r=1}^n \left (f(r) - f(r-1) \right) &= \cancel{f(1)} - f(0) + \cdots\\
&&&\quad\, \cancel{\f(2)}-\cancel{f(1)} + \cdots \\
&&&\quad\, \cancel{f(3)}-\cancel{f(2)} + \cdots \\
&&&\quad\, +\cdots + \cdots \\
&&&\quad\, \cancel{f(n-1)}-\cancel{f(n-2)} + \cdots \\
&&&\quad\, f(n)-\cancel{f(n-1)} \\
&&&=f(n) - f(0)
\end{align*}
- If \(f(r) = r^2(r+1)^2\) then
\begin{align*}
&& f(r) - f(r-1) &= r^2(r+1)^2 - (r-1)r^2 \\
&&&= r^2((r+1)^2-(r-1)^2) \\
&&&=4r^3
\end{align*}
Therefore \begin{align*}
&& \sum_{r=1}^n 4r^3 &= n^2(n+1)^2-0 \\
\Rightarrow && \sum_{r=1}^n r^3 &= \frac{n^2(n+1)^2}{4} \\
\end{align*}
- So
\begin{align*}
&& \sum_{r=1}^{2n+1} (-1)^{r-1}r^3 &= \sum_{r=1}^{2n+1} r^3 - 2 \sum_{r=1}^n (2r)^3 \\
&&&= \frac14(2n+1)^2(2n+2)^2 - 8 \frac{n^2(n+1)^2}{4} \\
&&&= (2n+1)^2(n+1)^2 - 2n^2(n+1)^2 \\
&&&= (n+1)^2(4n^2+4n+1 - 2n^2) \\
&&&= (n+1)^2(2n^2+4n+1)
\end{align*}
From the facts
\begin{alignat*}{2}
1 & \quad=\quad & & 0\\
2+3+4 & \quad=\quad & & 1+8\\
5+6+7+8+9 & \quad=\quad & & 8+27\\
10+11+12+13+14+15+16 & \quad=\quad & & 27+64
\end{alignat*}
guess a general law. Prove it.
Hence, or otherwise, prove that
\[
1^{3}+2^{3}+3^{3}+\cdots+N^{3}=\tfrac{1}{4}N^{2}(N+1)^{2}
\]
for every positive integer \(N\).
[Hint. You may assume that \(1+2+3+\cdots+n=\frac{1}{2}n(n+1)\).]
Show Solution
\begin{align*}
&& (n^2+1) + (n^2+2) + \cdots + (n+1)^2 &= n^3+(n+1)^3 \\
\Leftrightarrow && \sum_{i=n^2+1}^{(n+1)^2} i &= n^3 + (n+1)^3 \\
&& \sum_{i=n^2+1}^{(n+1)^2} i &= \sum_{i=1}^{(n+1)^2} i- \sum_{i=1}^{n^2} i \\
&&&= \frac{(n+1)^2((n+1)^2+1)}{2} - \frac{n^2(n^2+1)}{2} \\
&&&= \frac{(n+1)^2(n^2+2n+2) - n^2(n^2+1)}{2} \\
&&&= \frac{2(n+1)^3+n^2(n^2+2n+1) - n^2(n^2+1)}{2}\\
&&&= \frac{2(n+1)^3+2n^3 + n^2(n^2+1) - n^2(n^2+1)}{2}\\
&&&= (n+1)^3+n^3
\end{align*}
\begin{align*}
&& \sum_{i=1}^{N^2} i &=(0^3+1^3)+ (1^3+2^3)+(2^3+3^3) + \cdots + ((N-1)^3+N^3) \\
&&&= 2 \left (1^3+2^3 + 3^3 + \cdots + (N-1)^3 \right) + N^3 \\
\Rightarrow && \sum_{i=1}^N i^3 &= \frac12 \left ( N^3+ \sum_{i=1}^{N^2} i \right) \\
&&&= \frac12 \left ( N^3 + \frac{N^2(N^2+1)}{2} \right) \\
&&&= \frac{N^2(N^2+1)+2N^3}{4} \\
&&&= \frac{N^2(N^2+2N+1)}{4} \\
&&&= \frac{N^2(N+1)^2}{4} \\
\end{align*}
If \(\left|r\right|\neq1,\) show that
\[
1+r^{2}+r^{4}+\cdots+r^{2n}=\frac{1-r^{2n+2}}{1-r^{2}}\,.
\]
If \(r\neq1,\) find an expression for \(\mathrm{S}_{n}(r),\) where
\[
\mathrm{S}_{n}(r)=r+r^{2}+r^{4}+r^{5}+r^{7}+r^{8}+r^{10}+\cdots+r^{3n-1}.
\]
Show that, if \(\left|r\right|<1,\) then, as \(n\rightarrow\infty,\)
\[
\mathrm{S}_{n}(r)\rightarrow\frac{1}{1-r}-\frac{1}{1-r^{3}}\,.
\]
If \(\left|r\right|\neq1,\) find an expression for \(\mathrm{T}_{n}(r),\)
where
\[
\mathrm{T}_{n}(r)=1+r^{2}+r^{3}+r^{4}+r^{6}+r^{8}+r^{9}+r^{10}+r^{12}+r^{14}+r^{15}+r^{16}+\cdots+r^{6n}.
\]
If \(\left|r\right|<1,\) find the limit of \(\mathrm{T}_{n}(r)\) as
\(n\rightarrow\infty.\)
What happens to \(\mathrm{T}_{n}(r)\) as \(n\rightarrow\infty\) in the
three cases \(r>1,r=1\) and \(r=-1\)? In each case give reasons for
your answer.
Show Solution
\begin{align*}
&& S &= 1 + r^2 + r^4 + \cdots + r^{2n} \\
&& r^2S &= \quad \,\,\,\, r^2 + r^4 + \cdots+r^{2n}+r^{2n+2} \\
\Rightarrow && (1-r^2)S &= 1 - r^{2n+2} \\
\Rightarrow && S &= \frac{1-r^{2n+2}}{1-r^2}
\end{align*}
\begin{align*}
&& S_n(r) &= r + r^2 + r^4 + r^5 + r^7 + \cdots + r^{3n-1} \\
&&&= 1 + r + r^2 + \cdots + r^{3n} - (1 + r^3 + r^6 + r^{3n}) \\
&&&= \frac{1-r^{3n+1}}{1-r} - \frac{1-r^{3n+3}}{1-r^3} \\
\\
\Rightarrow && \lim_{n \to \infty} S_n(r) &= \frac{1-0}{1-r} - \frac{1-0}{1-r^3} = \frac{1}{1-r} - \frac{1}{1-r^3}
\end{align*}
\begin{align*}
&& T_n(r) &= 1 + r^2 + r^3 + r^4 + r^6 + \cdots + r^{6n} \\
&&&= \frac{1-r^{6n+6}}{1-r^6} + \frac{r^2-r^{6n+2}}{1-r^6} + \frac{r^3-r^{6n+3}}{1-r^6} + \frac{r^4-r^{6n+4}}{1-r^6} \\
&&&= \frac{1+r^2+r^3+r^4-r^{6n}(r^2+r^3+r^4+r^6))}{1-r^6} \\
\\
&&\lim_{n \to \infty} T_n(r) &= \frac{1+r^2+r^3+r^4}{1-r^6}
\end{align*}
If \(r > 1\) clear it diverges. if \(r = 1\) same story. if \(r = -1\) the sums in blocks of \(4\) are all \(1+1-1+1 = 2 > 0\) and so it also diverges.
The sequence \(a_{1},a_{2},\ldots,a_{n},\ldots\) forms an arithmetic progression. Establish a formula, involving \(n,\) \(a_{1}\) and \(a_{2}\) for the sum \(a_{1}+a_{2}+\cdots+a_{n}\) of the first \(n\) terms.
A sequence \(b_{1},b_{2},\ldots,b_{n},\ldots\) is called a double arithmetic progression if the sequence of differences
\[
b_{2}-b_{1},b_{3}-b_{2},\ldots,b_{n+1}-b_{n},\ldots
\]
is an arithmetic progression. Establish a formula, involving \(n,b_{1},b_{2}\) and \(b_{3}\), for the sum \(b_{1}+b_{2}+b_{3}+\cdots+b_{n}\) of the first \(n\) terms of such a progression.
A sequence \(c_{1},c_{2},\ldots,c_{n},\ldots\) is called a factorial progression if \(c_{n+1}-c_{n}=n!d\) for some non-zero \(d\) and every \(n\geqslant1\). Suppose \(1,b_{2},b_{3},\ldots\) is a double arithmetic progression, and also that \(b_{2},b_{4},b_{6}\) and \(220\) are the first four terms in a factorial progression. Find the sum \(1+b_{2}+b_{3}+\cdots+b_{n}.\)
Show Solution
Since the common difference is \(a_2 - a_1\) we can find that \(a_n = a_1 + (n-1)(a_2-a_1)\), then
\begin{align*}
&& a_1 &&+&& a_2 &&+&& \cdots &&+&& (a_1 + (n-2)(a_2 - a_1) && + && (a_1 + (n-1) (a_2 - a_1)) \\
+ && (a_1 + (n-1) (a_2 - a_1))&&+&& (a_1 + (n-2)(a_2 - a_1)&&+&& \cdots &&+&& a_2 && + && a_1 \\
\hline \\
= && 2a_1 + (n-1)(a_2 - a_1) && + && 2a_1 + (n-1)(a_2 - a_1) && + && \cdots && + 2a_1 + (n-1)(a_2 - a_1) && + 2a_1 + (n-1)(a_2 - a_1) \\
= && n(2a_1 + (n-1) (a_2 - a_1))
\end{align*}
Therefore the sum is \(a_1 n + \frac{n(n-1)}{2} (a_2 - a_1)\).
Since \(b_n - b_1 = (b_n - b_{n-1}) + (b_{n-1} - b_{n-2}) + \cdots (b_2 - b_1)\), \(b_n - b_1 = a_1 (n-1) + \frac{(n-1)(n-2)}{2}(a_2 - a_1) = (b_2-b_1)(n-1) + \frac{(n-1)(n-2)}{2}(b_3 -2b_2 +b_1)\). So \(b_n = b_1 + (b_2 - b_1)(n-1) + \frac{(n-1)(n-2)}{2} (b_3 - 2b_2 + b_1)\).
In particular
\begin{align*}
\sum_{i=1}^n b_i &= \sum_{i=1}^n \l b_1 + (b_2 - b_1)(n-1) + \frac{(n-1)(n-2)}{2} (b_3 - 2b_2 + b_1)\r \\
&= nb_1 + (b_2-b_1) \frac{n(n-1)}{2} + \frac{n(n-1)(n-2)}{6}(b_3-2b_2+b_1)
\end{align*}
Let \(b_2 - b_1 = x\) and \(b_3 - 2b_2+b_1 = y\), then
\begin{align*}
b_4 - b_2 &= d &= &2x + 3y \\
b_6-b_4 &= 2d &=& 2x +(10-3)y \\
&&=&2x + 7y \\
220-b_6&=6d &=& 220-(1 + 5x + 10y) \\
\end{align*}
\begin{align*}
&& 4x + 6y &= 2x + 7y \\
&& 6x+21y &= 219-5x-10y \\
\Rightarrow && 2x - y &= 0 \\
&& 11x + 31y &= 219 \\
\Rightarrow && x &= 3 \\
&& y &= 6
\end{align*}
Therefore the final sum is
\begin{align*}
n + 3 \frac{n(n-1)}{2} + 6 \frac{n(n-1)(n-2)}{6} &= n^3-\frac32n^2+\frac32n
\end{align*}
A firm of engineers obtains the right to dig and exploit an undersea tunnel. Each day the firm borrows enough money to pay for the day's digging, which costs £\(c,\) and to pay the daily interest of \(100k\%\) on the sum already borrowed. The tunnel takes \(T\) days to build, and, once finished, earns £\(d\) a day, all of which goes to pay the daily interest and repay the debt until it is fully paid. The financial transactions take place at the end of each day's work. Show that \(S_{n},\) the total amount borrowed by the end of day \(n\), is given by
\[
S_{n}=\frac{c[(1+k)^{n}-1]}{k}
\]
for \(n\leqslant T\).
Given that \(S_{T+m}>0,\) where \(m>0,\) express \(S_{T+m}\) in terms of \(c,d,k,T\) and \(m.\)
Show that, if \(d/c>(1+k)^{T}-1,\) the firm will eventually pay off the debt.
Show Solution
After \(n\) days they will have borrowed \(c\) for \(n-1\) days, \(c\) for \(n-2\) days, etc until \(c\) for no days.
Therefore the outstanding balance will be:
\begin{align*}
c + (1+k)\cdot c+ (1+k)^2 \cdot c + \cdots + (1+k)^{n-1} \cdot c &= c\frac{(1+k)^n-1}{(1+k)-1} \\
&= \frac{c[(1+k)^n-1]}{k}
\end{align*}
At the end of \(T\) days the outstanding balance will be \(S_T = \frac{c[(1+k)^T-1]}{k}\).
We can think of each payment of \(d\) during the subsequent period as being equivalent of a payment of \(d (1+k)^{m-1}\) \(m\) days later (as otherwise they would have accrued the equivalent amount in interest. Therefore after \(m\) days the amount paid back (equivalent) is:
\begin{align*}
(1+k)^{m-1} \cdot d + (1+k)^{m-2} \cdot d + \cdots + d &= \frac{d[(1+k)^m-1]}{k}
\end{align*}
Therefore the net position, \(S_{T+m}\) will be:
\begin{align*} S_{T+m} &= \frac{c[(1+k)^T-1](1+k)^m-d[(1+k)^m-1]}{k} \\
&= \frac{(1+k)^m [c ((1+k)^T-1)-d]+d}{k}
\end{align*}
Therefore they will eventually pay back their debts if \( [c ((1+k)^T-1)-d]\) is negative. ie \(d > c((1+k)^T-1) \Rightarrow d/c > (1+k)^T-1\)