Kamil N. GCSE Maths tutor, A Level Maths tutor, 13 plus  Maths tutor,...

Kamil N.

Currently unavailable: for new students

Degree: Mathematics and Computer Science (Masters) - Oxford, Merton College University

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About me

About me:

Hi, I'm Kamil - a first year Mathematics & Computer Science undergraduate from Oxford University. I am looking to tutor in my favourite subject – Mathematics.

Whether you would like some help with your homework, preparation for exams or mathematical competitions, or maybe you just want to enhance your mathematical skills, I would be more than happy to share with you my experience and enthusiasm in Maths. I can teach any topic from primary school up to high school.

I have many achievements in prestigious competitions such as Olympiads in Mathematics, Computer Science and Physics*. I also had excellent grades in my high school exams. Although I began tutoring quite recently, I have already had many satisfied students. Two of them needed a last-minute preparation for a resit and thanks to an intensive work with me they passed with very good grades.

The session:

I try to make my sessions enjoyable and challenging. My focus is always on understanding the ideas because this is what will make you succeed in the exams. I encourage my students to ask questions and to think independently. Apart from Maths, I can also explain to you a few topics in Physics and Computer Science or give some advice about applying to Oxford.

Other information:

* A full list of my achievements (most of them are Polish):

Finalist of 65th and 64th Polish Mathematical Olympiad (2012-2014)

Finalist of 63rd Polish Physics Olympiad (2013/2014)

Finalist with distinction of 20th Polish Olympiad in Computer Science (2012/2013)

Representing Oxford University in North-Western European Regional Contest 2014 - official regional contest in ACM International Collegiate Programming Contest with 26 place out of 96 teams

Final 10 of Electoral Gazette's competition “High school student of Wroclaw”

Laureate of Scientific League and 1st place as a team in mathematical sciences

2nd place in Lower Silesia Team Programming Championships

2nd place in an IT competition “Wielka Przesmycka”

Laureate of Polish Mathematical Olympiad for Secondary School Students

Laureate of a competition in Lower Silesia “Zdolny Slazak” in Mathematics and in Physics

1st place in Wroclaw Young Mathematicians Tournament

Winning Mathematical Marathon for primary schools

Winning Wroclaw Fast Calculation Championships for primary schools

Winning Lower Silesian Mathematical Matches for primary shools

Many prizes in Mathematical Kangaroo, Alfik, Polish Qualifications for the International Mathematics and Logic Games Championships

Two times junior champion of Lower Silesia in Chess

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Maths A Level £20 /hr
Maths GCSE £18 /hr
Maths 13 Plus £18 /hr

Qualifications

QualificationLevelGrade
MathematicsA-Level100% at basic and extended level
Computer ScienceA-Level100% at extended level
Physics and AstronomyA-Level100% at extended level
EnglishA-Levelwritten part - basic level 100%, extended level 97%; oral part – 100%
GermanA-Levelwritten part - basic level 100%
PolishA-Levelwritten part - basic level 84%; oral part 55%
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for new students

Questions Kamil has answered

What is the sum of the first n terms of a geometric sequence and where does it come from?

Recall that a geometric progression is a sequence (a_n) defined as follows   a_1 = a a_n = a * r^(n - 1) for all integers n > 1 where a and r are some fixed numbers and r /= 0 is called the common ratio of a sequence.   Let's denote S_n as the sum of the first n terms of this sequence.   ...

Recall that a geometric progression is a sequence (a_n) defined as follows

 

a_1 = a

a_n = a * r^(n - 1) for all integers n > 1

where a and r are some fixed numbers and r /= 0 is called the common ratio of a sequence.

 

Let's denote S_n as the sum of the first n terms of this sequence.

 

Case r = 1

Then all the terms are equal to a so

S_n = n*a

 

Case r /= 1

Then we have

S_n = a_1 + a_2 + … + a_n = a + a*r + … a*r^(n - 1) = a * (1 + r + … + r^(n – 1))

 

Now the formula for the difference between two n-th powers tells us that

r^n - 1 = r^n – 1^n = (r - 1) * (1 + r + … + r^(n-1))

and since r /= 1 we can divide both sides by r – 1 to have

1 + r + … + r ^(n - 1) = (r^n - 1) / (r – 1)

 

Finally, substituting this expression into the first equality we get

S_n = a * (r^n - 1) / (r – 1)

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2 years ago

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