I'm a second year student at Exeter University. I've been passionate about maths from a young age and have carried that love through my school and uni career. I hope that I can generate the same passion for my tutees as well.
I'm a patient and friendly person, and have always enjoyed helping friends and family with their work. I enjoy teaching people about my subject and I will always bring enthusiasm to our sessions.
You guide the topics that we will cover in our sessions. I'll make sure you understand the basics before we tackle the harder topics, once you have basic understanding of a topic, the rest falls into place.
I will do my best to find out what style of learner you are so that I can tailor each tutorial to you, such as including more examples, more diagrams or by finding new ways of explaining ideas. I will not simply talk at you, I will be engaging and make sure that any question you have is answered.
If you have any questions, don't hesitate, send me a "WebMail" or organise a Free Trial Session, I look forward to meeting you!
|Maths||A Level||£20 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Ami (Parent) September 10 2016
Fergus (Parent) May 18 2016
Lucy (Parent) April 28 2016
Danielle (Parent) April 19 2016
To differentiate Xa, where a is any real number, you multiply X by a, and subtract 1 from the power.
First you must show that the statement on the right hand side is true for n=1:
Σi=0 i2 when n=1, is equal to 12=1
This means that the statement is true for n=1.
Next you assume that it is true for 'k', where k is any number, and so you get;
Σi=0 i2 when n=k, is equal to (k/6)(k+1)(2k+2)
You then have to show that the statement is true for n=k+1 which would make;
Σi=0 i2 when n=k+1, is equal to (k+1)/6(k+2)(2k+3) call this 1)
As the left hand side is a sum, it can be written as;
Σi=0 i2 when n=k + (k+1)2
We already know the sum of i2 when n=k and so we can substitute it in;
(k/6)(k+1)(2k+1) + (k+1)2
We then try and reach 1)
We can factorise out (k+1)
Next, multiply the inner brackets;
Take out a factor of 1/6
Finally, factorise the inner bracket;
As this is equal to 1), we have proven that the statement is true for all values of n.see more
Standard form is a way of writing large numbers to make them more compact, for example
Instead of writing:
197,000,000= 1.97 x 100,000,000
And so we can write:
1.97 x 108, which is more concise.see more