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XICHEN
- Rate £11
- Response 4h
-
Students3
Number of students XICHEN has accompanied since arriving at Superprof
Number of students XICHEN has accompanied since arriving at Superprof
£11/hr
See Maths tutors
Unfortunately, this tutor is unavailable
- Maths
- Algebra
- Geometry
- Statistics
Mathematics student offering maths (including Geometry, Algebra, Statistics etc.) lessons up to first year university level in Bristol
- Maths
- Algebra
- Geometry
- Statistics
Lesson location
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At XICHEN's house: Bristol
- Online
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About XICHEN
I am a second year student in University of Bristol majoring in maths. Two and a half years of university study has given me good mathematical derivation and computing skills. In the summer of 2018, as a maths tutor, in a half-year part-time teaching, I helped a boy with difficulty learning mathematics effectively improve his mathematical ability and made him interested in mathematics. In this teaching, I used some examples from real life to explain abstract mathematical problems (such as the relationship between geometry and algebra), and cooperated with certain homework exercises and encouragement, step by step, finally effectively improved his confidence in learning mathematics.
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About the lesson
- Primary
- Secondary
- GCSE
- +8
levels :
Primary
Secondary
GCSE
AS Level
A Level
BTEC
Adult Education
Undergraduate
Masters
Diploma
Doctorate
- English
All languages in which the lesson is available :
English
Feel relaxed about maths! Sometimes you feel puzzled because you think mathematical problems in a bit complicated way. A tricky example (for junior high school students): How to solve [1/(1*2)]+[1/(2*3)]+...+[1/(99*100)]. Of course you can compute it directly but it is wiser to do in this way: Since 1/(1*2)=(1/1)-(1/2), 1/(2*3)=(1/2)-(1/3), ... ,1/(99*100)=(1/99)-(1/100), the previous fomula = (1/1)-(1/2)+(1/2)-(1/3)+...+(1/98)-(1/99)+(1/99)-(1/100) = 1-(1/100)=99/100. See! You have solved it. For mathematics, I will teach you how to find "the most general description" of a maths problem. Please think about this question: how to solve k/[m*(m+n)]+k/[(m+n)*(m+2n)]+...+k/[(m+(t-1)n)*(m+tn)], where t is a positive integer and all k,m,n are rational numbers! (This is what I said "the most general description")
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Rates
Rate
- £11
Pack prices
- 5h: £55
- 10h: £110
online
- £11/h
Details
I don't need tariffs but if possible, I recommend students to come to my college dormitory since it is better to learn maths face to face.
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