Thank you SO MUCH for the important part you played in helping Adam get the A level grades he needed to get to the course of his choice in Newcastle. To anyone contemplating getting Adams support - just do it :-) - having a helpful, knowledgable, friendly, same sort of age chap to ask about topics you need a bit more time on to get clarity - is invaluable. And Adam is reliable - flexible - friendly and supportive. Thanks and best of luck. Brenda - Mum to the Adam :-) - this Adam tutored.

Brenda, Parent from berks

Why waste time looking locally when itâ€™s easier to find the right tutor online?

Tutoring is easier and more flexible when you remove the need to plan around travel

With live online one-to-one sessions you’re always engaged

Your live sessions are recorded, so you can play tutorials back if you want to revise

Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after lessons. In my tutorials, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

Proof by induction has three core elements to it. To start with you must prove that the statement is true for the 'basic case'. For the most part this is 1, but some questions state it is higher.

Do this by subbing 1 into the equation and ensuring that it is divisible by 3

1^3 =1

5(1)=5

1+5=6 6/3=2 Therefore divisible by three and true for 1.

Then in order to futher prove it, we are going to assume that this is true for n=k

leaving us with the equation k^3+5k=3a as it is divisible by 3.

The next stage is to prove true for n=k+1.

Do this by subbing k+1 into the original equation:

(k+1)^3 +5(k+1)

multiplying this out gives:

k^3+3k^2+3k+1+5k+5

Now we have already established that k^3+5k=3a so through rearranging, k^3=3a-5k.

Subbing this into the k+1 equation gives us:

3d+3k^2+3k-6. Each element is a multiple of three so by taking three out leaves us:

3(d+k^2+k-2) which is a multiple of three and thus divisible by three.

Then leave a concluding statement along the lines of:

'As n^3+5n is true for n=k, then it is true for n=k+1. As it is true for n=1, then it must be true for n is greater than 1'

Answered by Philip D.

Studies Mathematics at Exeter

When dealing with powers of complex numbers, always start by putting the quantity into exponential form.

** i** has a magnitude of

**i = exp(iπ/2)**

Now the expression is in exponential form, taking the square root is easy, using basic exponential math.

**sqrt(i) = (exp(iπ/2))^(1/2) = exp(iπ/4)**

This quantity has a modulus of** 1** and an argument of

**sqrt(i) = (1 + i)/sqrt(2)**

let y=arsinh(x)

x=sinh(y)=(e^{y} - e^{-y})/2

2x=e^{y }- e^{-}^{y}

2x*e^{y}=e^{y}-1 (multiply bye^{y})

0=(e^{y})^{2}-2xe^{y}-1

This is a quadratic in e^{y} with coefficients: a=1,b=-2x,c=-1

Usinng the quadratic formula (and simplifying):

e^y=x +/- sqrt(x^{2}+1)

but e^{y}=x-sqrt(x^{2}+1) isn't possible as e^{y}>0 for all y.

so e^{y}=x+sqrt(x^{2}+1)

y=ln(x+sqrt(x^{2}+1))

arsinh(x)=ln(x+sqrt(x^{2}+1)).

(Note that sqrt(x) is standard notation for 'the square root of x' on computers).

This is a typical further maths question, doing it correctly is a matter of carrying out a two-step process.

Start by finding the determinant of the matrix,

det(A)=ad-bc

Then swap the entries a d and negate the other entries. After dividing by the determinant the inverse of A is given.

A^-1=1/(ad-bc)([d -b],[-c, a]).

When it comes to answering questions about inequalities, it is important to remember the signs and what they represent. In this instance, we need to find a range of solutions where 2(3x+1) is **greater than **3-4x.

To solve this inequality, we need to make x the subject of the inequality. First, we need to expand 2(3x+1) to get 6x+2. Now we have the inequality 6x+2>3-4x. Next we rearrange to make x the subject. By adding 4x to both sides and subtracting 2 from both sides, we get the inequality 10x>1. Finally, we divide both sides by 10 to get x by itself. The simplified inequality is x>1/10. Therefore the answer to the question is the range of solutions for the inequality 2(3x+1)>3-4x is x is **greater than** 1/10.

Answered by Samradnyee K.

Studies Biomedical Engineering at Kings, London

Induction is an incredibly powerful form of proof. As such, it is essential to be able to call upon such a tool if you want to study any higher form of mathematics.

The basic structure of the method is as follows:

1. Show that the claim is true for a base case (usually n=0 or n=1, but it will work for higher numbers)

2. Assume that the claim holds true for some arbitrary value (we usually call this the n=k^{th} case)

3. Prove that, given step 2, it holds true for the n=k+1^{th} case

The idea of this is that if it is true for n=1, and you’ve shown that if it is true for n=some number then it is true for n=that number +1, then it must be true for n=2. If it is true for n=2, then it must be true for n=3 ==> true for n=4 ==> true for n=5 ==>… etc

Let’s work through an example: Show that the sum of the first n squares is equal to n(n+1)(2n+1)/6

1. Let n=1. The sum of the first n squares= the sum of the first 1 square=1^{2}=1. Now note that (1)(1+1)((2)(1)+1)/6 = (2)(3)/6 = 1. So our formula works for this particular case- it is true for the case that n=1.

2. Assume that the sum of the first k squares is k(k+1)(2k+1)/6

3. Consider the sum of the first k+1 squares. This is equal to the sum of the first k squares plus (k+1)^{2}. Using step 2, we have that the sum of the first k squares is k(k+1)(2k+1)/6, so the sum of the first k+1 squares is (k(k+1)(2k+1)/6) + (k+1)^{2} = (k+1)((k(2k+1)/6)+(k+1))=(k+1)(k(2k+1)+6k+6)/6 = (k+1)(2k^{2}+7k+6)/6 = (k+1)(2k+3)(k+2)/6. Note that this can now be written as (k+1)((k+1)+1)(2(k+1)+1)/6, which is in the form of n(n+1)(2n+1)/6 where n=k+1, so assuming step 2, we’ve proved that it is true for n=k+1.

Finally, since this formula is true for n=1, and given that we’ve shown that if it is true for n=k, then it is true for n=k+1, then it must be true for all integer values of n greater than or equal to 1. So, the sum of the first n squares is equal to n(n+1)(2n+1)/6.

All the induction proofs you’ll be asked to do follow this structure. The only difference between them is the algebraic manipulations.

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OurÂ FurtherÂ Maths A Level TutorsÂ will give you that focused, individual attention which could make all the difference to your grade. Gaps in your learning, careless mistakes and ineffective time-management during the exams can all be overcome with the expert guidance of aÂ FurtherÂ Maths A-Level Tutor.Â YourÂ Maths A Level TutorÂ will also ensure you have plenty of past papers to complete, as practice, repetition and excellent exam technique are key to Maths A Level success.