Degree: Theoretical Physics with Applied Maths (Masters) - Birmingham University
I myself am studying Theoretical Physics with Applied Maths in the University of Birmingham, and I would absolutely love tutoring (especially Maths!).
I am very patient and motivated - since I know that understanding some concepts can be daunting at GCSE and A-level.
I aim to understand the student's position whilst approaching concepts in a logical manner at an appropriate pace. Provided I am informed in advance of the topic for an arranged lesson, I will give my time to prepare fully around it before the sessions. And/Or if there are any questions or past-paper help I will guide through the important facts through the paper.
In any case, more importantly, I believe the tutee him/herself should lead the sessions, if they are going to be effective.
Normally in the sessions I would give explanations along with worked examples discussing with the tutee.
In the midst I would ask reasonably progressive questions to make sure the tutee can understand the relevant concepts.
I also achieved Silver Award in the Maths Challenge in my year 12.
Availability: Available most evenings except Monday evenings
|Further Mathematics||A Level||£24 /hr|
|Maths||A Level||£24 /hr|
|Physics||A Level||£24 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Luke (Student) October 26 2015
Sheeru (Student) April 26 2015
Iftikhar (Parent) March 31 2015
Sheeru (Student) April 4 2015
Any geometric series is defined by an initial term a, and a common ratio b.
This means that we start with a and multiply by b to get each next term.
And so the general geometric series is written:
a*r^0,a*r^1,a*r^2,a*r^3,...,a*r^n ; where n+1 = no. of terms.
Now the sum of the above, S is:
S = a*r^0+a*r^1+a*r^2+a*r^3+...+ar^n
Factorise r out in most terms in right hand side:
S = a*r^0+r*(a*r^0+a*r^1+a*r^2+...+a*r^(n-1))
Now note that the sum within the brackets includes all terms in our original sum S, save the last term a*r^n. This means we can substitute (S-a*r^n) for it. Therefore,
S = a*r^0+r*(S-a*r^n)
Now all we are left to do is make the value of the sum, S, the subject of the equation:
S = a+r*S-a*r^(n+1)
S-r*S = a-a*r^(n+1)
S*(1-r) = a*(1-r*(n+1)); factorize S on LHS, and a on RHS.
S = a*(1-r*(n+1))/(1-r)