Q1
A student claims that the following proposition is true for all positive integers n.
1 *1+ 3 * 2 + 3^2*3+…+3n-1*n=3^n(2n-1)+1/4
(a)By using Mathematical Induction, confirm the student’s claim.
(b)Can the student include the case n = 0 in his proposition? Explain.
Q2
(a) Each of the two boys (Alex and Marvin) and three girls (Candy, Ellen and Venus) scored a
different grade in a recent Mathematics test. The students made the following observations on
their results (namely, A, B, C, D and E, A being the best and E being the worst):
(O1) Alex did better than two or more of the girls;
(O2) The worst grade was not scored by a girl;
(O3) Venus did not score an A;
(O4) Candy scored a C.
[Note: In answering the following questions, you may like to make use of the following
template, with student name(s) or initial(s) on each scoring line.
(i) If Alex did not score an A, what were the grades scored by Alex, Ellen and Venus?
Explain Briefly.
A _____________________
B ______________________
C ______________________
D ______________________
E ______________________
(ii) If a boy scored an A, who was this boy? What were the possible grades scored by Ellen
and Venus? Explain Briefly.
(iii) If all the girls did better than the boys, which of the four observations must be incorrect?
Explain briefly.
(b) Catherine and Christine take turns breaking a piece of chocolate consisting 5 x 10 small
square pieces. At each turn, they may only break along the division lines of the small square
pieces and it is agreed that Catherine will start the breaking first.
(i) If they keep breaking the pieces up until only the small square pieces remain and no more
breaking up is possible, who will win the game? Explain briefly.
(ii) If the winner is the one who first obtains a single piece of the small square pieces, who will win the game and what will be her winning strategy?
[急]MI及其他數學問題
2014-10-30 4:59 pm
回答 (1)
2014-10-31 3:23 am
✔ 最佳答案
Q1:(a) 1 *1+ 3 * 2 + 3^2*3+…+3^(n-1)*n=[(3^n)(2n-1)+1]/4 for all positive integer n
When n=1
LHS=1 *1=1
RHS=[3^1(2*1-1)+1]/4=1
Hence the proposition is true for n=1.
Assume the proposition is true for n=k.
By adding the (k+1)th term onto both sides,
LHS=1 *1+ 3 * 2 + 3^2*3+…+3^(k-1)*k+3^[(k+1)-1]*(k+1)
RHS=[(3^k)(2k-1)+1]/4 +3^ [(k+1)-1]*(k+1)
=[(3^k)(2k-1)+1]/4 +(3^k)(k+1)
=[(3^k)(2k-1)+1+4(3^k)(k+1)]/4
=[(3^k)(2k-1+4k+4)+1]/4
=[(3^k)(6k+3)+1]/4
=[(3^k)3(2k+1)+1]/4
={[3^(k+1)] [(2k+2-1] +1}/4
={[3^(k+1)] [2(k+1)-1]+1}/4
The proposition can be induced for n=k+1.
Hence the proposition is true for all positive integer n.
(b) The student can include the case n = 0 in his proposition.
When n=0, LHS=0, RHS=[(3^0)(2*0-1)+1]/4=0=LHS
He can redefined the range of n as: for all non-negative integer n.
2014-10-30 19:26:16 補充：
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