数学Paper 1讨论专区

walrein_lim88

回复 800# bushman_tong


    LOL ..我不是老师。。。才考完STPM吧了。。

bushman_tong

小弟在这儿想请教各位大哥大姐..

1)Given that n is a positive integer,
(a) Find in tern of S1, the sum of integers from 2n to 4n inclusive.
(b) Find in term of S2, the sum of odd integers from 2n to 4n.

2)Positive integers are grouped in brackets:

   (1) , (2,3) , (4,5,6) ,.........with r integers in the rth bracket.

(a) Find the expression for the first and last term in the rth bracket.
(b) Find the sum of all the integers in the first 20 brackets.
(c) Prove that the sum of integers in the rth bracket is (r/2)[(r^2)+1]

希望各位可以帮到我..谢谢!

bushman_tong

回复 801# walrein_lim88



不好意思...误会了...

walrein_lim88

回复 802# bushman_tong

walrein_lim88

回复 802# bushman_tong


   

bushman_tong

回复 805# walrein_lim88


    小弟有不明之处...

   如何找 An=1+(n/2)(1+n)    ?

   还有为什么last term in the rth bracket 要加 (r - 1)(1)?

walrein_lim88

回复 806# bushman_tong


    对不起。。我打错。。是Ar=1+0.5(r-1)(r)
   解释：
   注意下1st term for each bracket, there is a pattern:
   1,    2,   4,    7,    11,    16....
     +1  +2   +3  +4      +5..........
   
   1st term: 1
   2nd term is: 1 + (1) -is summation of AP
   3rd term is : 1+  (1+2) summation of AP - look at the last term : r-1(3-1=2)
   4th term is : 1 + (1+2+3) summation of AP look at last term: r-1 (4-1=3)

  deduce that:
for 1st term for rth bracket:
  1+ (1+2+3+..+r-1)
Use AP summation formula:
1+ 0.5(r-1)(r-1+1)=1+0.5(r-1)(r)


  

walrein_lim88

回复 806# bushman_tong


     还有为什么last term in the rth bracket 要加 (r - 1)(1)?
    现在我们知道。。for every 1st term in rth bracket:
     =1+(0.5)(r-1)(r)
     我们看下LAST TERM：
     注意上面的PATTERN：
    (1), (2,3),(4,5,6)
    1st bracket = 1 term
    2nd bracket = 2 term
    3rd bracket = 3 terms
    so..if rth bracket = got r terms
然后再注意下里面的号码，其实也是AP ，where d = 1
   so in rth bracket, a=1+0.5(r-1)(r)  and d = 1 and there are r terms totally
so the last term Tr= a+ (r-1)d
                       Tr=1+0.5(r-1)(r)+(r-1)(1)

bushman_tong

回复 808# walrein_lim88


    好难理解的题目..


   谢谢!

walrein_lim88

回复 809# bushman_tong


    你明白吗？？

bushman_tong

回复 810# walrein_lim88


勉强可以...

还需要更深入的思考...

bushman_tong

我快崩溃了...sequences and series 很难阿!!!

silent91

我快崩溃了...sequences and series 很难阿!!!
bushman_tong 发表于 13-3-2010 08:19 PM

中六好像没有东西是容易的咯～

bushman_tong

回复 813# silent91


    可以这么说...不然很多人就不用酱辛苦了...

walrein_lim88

回复 814# bushman_tong


    还是不明白吗？？？

bushman_tong

回复 815# walrein_lim88


    半懂半不懂....

    我还没到你所思考的境界...

    现在做着sequences and series 也是要死要活...

bushman_tong

这题也是伤脑筋...


           n
Find  ∑ (n+1+2r)  in term of n.
         r=0

芭樂

回复 817# bushman_tong

r = 0 > n+1
           n
=n+1 + ∑ (n+1+2r)
           1
           n     n   n
=n+1 + ∑ n+∑1+2∑r)
           1     1    1
                    n
=n+1+n^2+n+2∑ (r)
                    1
=n^2+n+1+2(n)(n+1)/2
=n^2+n+1+n^2+n
=2n^2+2n+1

芭樂

我校的math paper 1
第一课到coordinate geometry
答案要假期后才能给

bushman_tong

回复 818# 芭樂


    可是答案不是这个...