I've some maths problems i cant solve and hope someone can help me.
1) For each of the following relations on the
set {1, 2, 3, 4},
decide whether it is reflexive,
symmetric, anti-symmetric and/or transitive.
i) {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3),
(4, 4)}
ii) {(1, 3), (1, 4), (2, 3), (3, 4)}
iii) {(1, 1), (1, 3), (1, 4), (2, 1), (2, 3),
(2, 4), (3, 1), (3,
3), (3, 4)}
2) What is the coefficient of x^2 y^7 in (x + y)^9 ?
3) Prove that 12+32+52+…+ (2n+1)2 = (n+1) (2n+1) (2n+3)/3 whenever n is a nonnegative integer.
Thanks again!
JC maths?
Originally posted by -StarDust-:JC maths?
Nope, computing maths. Dip IT course
Hi,
For Q1, check this out:
http://mathforum.org/library/drmath/view/54256.html
For Q2, pick up any maths reference book to read about binomial expansions.
For Q3, 12 + 32 + 52 + ... + nth term looks like an arithmetic series whose first term is 12 and common difference is 20. I wonder if you have typed the series correctly...
Thanks.
Cheers,
Wen Shih
1(i) is reflexive since (For all x, x is related to x). 1(i) is also symmetric (i.e., (x,y) \in R => (y,x) \in R). Should not be hard to check the definition for the rest of Q1
Q2 Try Googling for Binomial Theorem
Q3 Can't make sense of this. E.g., when n=1, LHS and RHS don't match up, i.e., LHS is 12, RHS = (2)(3)(5)/3 = 10. Maybe you can explain the notation more, since on the LHS what you meant by "(2n+1)2" would make more sense as "(2n+1)*10 + 2"