Then

{jatex options:inline}\begin{equation} \begin{aligned} a_{n+1}+b_{n+1} \sqrt{5} &= (2+3 \sqrt{5} )^{n+1} \\ &=(2+3 \sqrt{5} )(2+3 \sqrt{5} )^n \\ &=(2a_n+15b_n)+(3a_n+2b_n)\sqrt{5} \end{aligned} \end{equation}{/jatex}.

We can then define {jatex options:inline}(2+3

{jatex options:inline}a_{n+1}=2a_n+15b_n, \: b_{n+1}=3a_n+2b_n{/jatex}. ]]>

We can simplify each term, obtaining the sequence

{jatex options:inline}\frac{1-sin^2 \theta}{sin \theta} =\frac{cos^2 \theta}{sin \theta} , \: cos \theta , \: sin \theta , \: \frac{1-cos^2 \theta}{cos \theta}=\frac{sin^2 \theta}{cos \theta} ,...{/jatex}

This is a geometric sequence with first term {jatex options:inline}a=\frac{cos^2 \theta}{sin \theta}{/jatex} and common ratio {jatex options:inline}r=\frac{sin \theta}{cos \theta}{/jatex}.

The {jatex options:inline}n^{th}{/jatex} term is {jatex options:inline}a_N=ar^{n-1}=\frac{cos^2 \theta}{sin \theta} (\frac{sin \theta}{cos \theta})^n{/jatex} and sum {jatex options:inline}S=\frac{a}{1-r}=\frac{cos^2 \theta / sin \theta }{1- sin \theta / cos \theta}=\frac{cos^3 \theta}{sin \theta cos \theta - sin^2 \theta}{/jatex}.

The expression for the sum of the series is only valid for {jatex options:inline}r=\frac{sin \theta}{cos \theta } = tan \theta \lt 1 \rightarrow \theta \lt \frac{\pi}{4}{/jatex}.]]>

n |
{jatex options:inline}(1 + 1/n)^n{/jatex} |

1 |
2 |

2 |
2.25 |

3 |
2.370370370 |

4 |
2.44140625 |

10 |
2.593742460 |

100 |
2.704813829 |

1000 |
2.716923932 |

10000 |
2.718145927 |

100000 |
2.718268237 |

The table shows that asgets largergets closer toWe can write

Writewhereis a fixed number.

so that as n tends to infinityThe solution to this differential equation isso putto obtain

]]>We can find a better estimate by writing {jatex options:inline}\sqrt{1300}=36+x{/jatex} where {jatex options:inline}x{/jatex} is small.

Then {jatex options:inline}1300=(36+x)^2=1296+72x+x^2{/jatex} (1)

Since {jatex options:inline}x{/jatex} is small, {jatex options:inline}x^2{/jatex} is very small and we can ignore it, so

{jatex options:inline}1300 \simeq 1296+72x{/jatex}.

{jatex options:inline}4 \simeq 72x{/jatex}.

{jatex options:inline}x \simeq \frac{4}{72} =0.0555555...{/jatex}.

Then {jatex options:inline}\sqrt{1300} \simeq 36.055555...{/jatex}.

We can improve this estimate by taking {jatex options:inline}x^2{/jatex} into account. From (1) we have {jatex options:inline}x^2+72x-4=0 \rightarrow x=\frac{4-x^2}{72}{/jatex}.

We can use this as an iteration expression {jatex options:inline}x_{n+1}=\frac{4-x_n^2}{72}{/jatex}, and by putting {jatex options:inline}x_1=\frac{1}{18}{/jatex} on the right hand side we get {jatex options:inline}x_2=\frac{4-(1/18)^2}{72} \simeq 0.055513{/jatex}, so that {jatex options:inline}\sqrt{1300} \simeq 37.055513{/jatex}.

This is correct to 8 significant figures. ]]>

Successive terms in an arithmetic sequence are defined bywhereis the common difference. We addto each term to get the next term. This means thatso that

Any term is the arithmetic mean of the term immediately preceding and the term immediately succeeding that term.

We can generalise this to any odd number of termsThe middle term is the arithmetic mean of the terms immediately preceding and the terms immediately succeeding:

In turn this can be generalized in the obvious way to an even number of terms.

Successive terms in an geometric sequence are defined bywhereis the common ratio. We multiply each term byto get the next term. This means thatandso that

From this we obtain

This equation says that any term is the geometric mean of the term immediately preceding and the term immediately succeeding that term.

We can generalise this to any odd number of terms,The middle term is the geometric mean of the terms immediately preceding and the terms immediately succeeding:

In turn this can be generalized in the obvious way to an even number of terms.

In fact a sequence can be any list ofnumbers. The arithmetic mean is just the sum of all the terms divided by the number of terms and the geometric mean is just the nth root of the product of the terms.

]]>The problem is to find the first term and the common differences.

We need to find an equation for the common difference, which is equal to the 2nd term minus the first term, or the 3rd term minus the second term.

{jatex options:inline}8k-7)-(3k-5)=(3k-5)-(k-2){/jatex}

{jatex options:inline}5k-2=2k-3{/jatex}

{jatex options:inline}5k-2k=2-3{/jatex}

{jatex options:inline}3k=-1 \rightarrow k=- \frac{1}{3}{/jatex}

The 1st term is {jatex options:inline}k-2=- \frac{1}{3}-2= - \frac{7}{3}{/jatex}

The common difference is {jatex options:inline}2k-2 = 2 \times - \frac{1}{3}- 3 = - \frac{11}{3}{/jatex}

An arithmetic sequence has third term 5 and seventh term 13. Find the sum of the first ten terms.

The formula for the nth term is {jatex options:inline}a_n=a+(n-1)d{/jatex} where {jatex options:inline}a, \: d{/jatex} are the first term and the difference of successive terms respectively. Hence

{jatex options:inline}a_3=5=a+(3-1)d=a+2d \rightarrow 5=a+2d{/jatex} (1)

{jatex options:inline}a_7=5=a+(7-1)d=a+6d \rightarrow 13=a+6d{/jatex} (2)

(2)-(1) gives {jatex options:inline}8=4d \rightarrow d=2{/jatex}

From (1) then {jatex options:inline}5=a+2 \times 2=a+4 \rightarrow a=1{/jatex}.]]>

4, 9, 14, 19, 24 is an arithmetic sequence because we add 5 to each term to get the next term. The general form for the nth term in a geometric sequence is:

whereis the first term andis the difference between any two successive terms.

Thefactor reflects the fact that to get the 1 ^{st } term we don't have to add anything: only from the 1 ^{st } term do we start adding things.

When we add up n terms, we write down an expression like,

By writing this backwards we obtain,

We can now add the two sequences, gettingon the left hand side and altogether n terms all the same,on the right hand side, so

Example: The 3 ^{rd } term of an arithmetic sequence is 9 and the 5 ^{th } term is 17. Find the first term, the common difference and the smallest value ofsuch that

and

Now solve the simultaneous equations

(1)

(2)

Sub into (1)

Solve

Non integer or negative values of n are not allowed here, because we are considering only the natural numbers, so

]]>It is quite tricky to calculate the month repayment. The method is illustrated in this example.

A loan of £500 is taken out at 11% annual interest to be repaid at a fixed monthly rate. Interest is to be charged from the date of the loan and the first repayment is to be made a month after the loan is taken out. Find the monthly repayment.

Call the monthly repayment x. Since the annual rate of interest is 11%, the monthly rate of interest is the solution to

After a month the amount owing in £ isand after the first monthly payment the amount owing is

After two months the amount owing in £ isand after the second monthly payment the amount owing is

After three months the amount owing in £ isand after the second monthly payment the amount owing is

We will carry on in the same way, until after two years, twenty four monthly repayments, the whole loan is paid off. After the last monthly payment there is no money left owing so

The expression on the right is a geometric series with first termand common ratioso the right hand side equalsand

Hence

Substituting the value ofinto this expression gives a monthly repayment of £23.18 to the nearest penny. The total amount repaid is £556.41 to the nearest penny.

]]>The calculations go like this for an interest rate of 10%:

£1000 deposited at start of year 1 earns 10% of £1000 in interest=

Amount in account at end of year 1=

Amount in account at start of year 2 =£1100

£1100 earns 10% of £1000 in interest=

Amount in account at end of year 2=

Amount in account at start of year 3 =£1210

£1210 earns 10% of £1210 in interest=

Amount in account at end of year 3=

Amount in account at start of year 4 =£1331

£1331 earns 10% of £1331 in interest= .

Amount in account at end of year 4=

And so on. It is important to be methodical and work out an answer for each year separately. Alternatively we could use the formula

The formula used for the example above would have given the answer

as above.

]]>{jatex options:inline}d= k- \frac{1}{k}, \: 2d=k^2+1-k \rightarrow 2k-\frac{2}{k}=k^2+1-k \rightarrow k^3-3k^2+k+2=0{/jatex}. This cubic expression factorises and we get {jatex options:inline}(k-2)(k^2-k-1)=0{/jatex}.

Since {jatex options:inline}k \in \mathbb{Q}{/jatex}, {jatex options:inline}k=2{/jatex} and the common difference is {jatex options:inline}d=k-\frac{1}{k}=2- \frac{1}{2}= \frac{3}{2}{/jatex}.

The first six terms are {jatex options:inline}- \frac{5}{2}, \: -1, \: \frac{1}{2}, \: 2, \: \frac{7}{2}, \: 5{/jatex}]]>