Math 213 Exam 1 (Solutions)
Prof. I.Kapovich February 22, 2008
Problem 1[20 points]
For each of the following statements indicate whether it is true or false.
You do not have to explain your answers.
(1) For the set S={∅,{∅},{{∅}}} we have |S|= 1.
(2) If |A|=mand |B|=n, then the number of all functions from Ato
Bis mn.
(3) For every n≥r≥1 we have C(n, r)≤P(n, r).
(4) Whenever E, F are events such that p(E) = 0, then Eand Fare
independent.
(5) For every n≥1 the number Pn
i=0 n
iis even.
Solution.
(1) FALSE. In fact, we have |S|= 3.
(2) FALSE. In fact, he number of all functions from Ato Bis nm.
(3) TRUE. Indeed, C(n, r) = P(n, r)/r! and so C(n, r )≤P(n, r).
(4) TRUE. Indeed, in this case p(E) = 0 and 0 ≤p(E∩F)≤p(E) = 0,
so that p(E∩F) = 0. Thus 0 = 0·p(F), that is p(E∩F) = p(E)p(F).
(5) TRUE. Indeed, Pn
i=0 n
i= 2nis even.
Problem 2[20 points] Prove that for every integer n≥1 we have:
12−22+ 32− · · · + (−1)n−1n2= (−1)n−1n(n+ 1)
2.
Give all the details of your work.
Solution.
We will prove this statement by induction.
Base of Induction.
For n= 1 we have 12= 1 = 1(1+1)
2, as required.
Inductive Step.
Suppose that n≥1 and that it is known that
(∗) 12−22+ 32− · · · + (−1)n−1n2= (−1)n−1n(n+ 1)
2
We need to show that
12−22+ 32− · · · + (−1)n−1n2+ (−1)n(n+ 1)2= (−1)n(n+ 1)(n+ 2)
2.
By adding (−1)n(n+ 1)2to both sides of (∗) we get:
12−22+ 32− · · · + (−1)n−1n2+ (−1)n(n+ 1)2= (−1)n−1n(n+ 1)
2+ (−1)n(n+ 1)2=
(−1)n(n+ 1) −n
2+ (n+ 1)= (−1)n(n+ 1)2n+ 2 −n
2= (−1)n(n+ 1)(n+ 2)
2,
as required.
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