数学Paper 1讨论专区

kelfaru

都说这不是普通的公式...
Allmaths 发表于 25-9-2010 04:55 PM

酱以你的人生经验,
integrate 爱 dx = ???

Allmaths

酱以你的人生经验,
integrate 爱 dx = ???
kelfaru 发表于 25-9-2010 04:59 PM

对不起我还没到那个境界...

hongji

1/3.5.7+1/5.7.9+1/7.9.11 ......
= (1/3)(1/5)(1/7)+(1/5)(1/7)(1/9)+(1/7)(1/9)(1/11)....
(1/3) , 1/5 ...
peaceboy 发表于 23-9-2010 12:26 AM

  真的谢谢阿

hongji

(3-x/2)^5 in ascending powers of 1/x
=(3/x-x^2)^5   <---这样吗？
那么
(3-x/2)^5 in ascending power of x
=(3(1-x/6)^5    ?

peaceboy

(3-x/2)^5 in ascending powers of 1/x
=(3/x-x^2)^5
hongji 发表于 25-9-2010 05:17 PM

    (3-x/2)^5=[(-x/2)(1-6/x)]^5
                  =(-x/2)^5 (1-6/x)^5  ,l-6xl<1



(3-x/2)^5 = [(3)(1-x/6)]^5
               = (3)^5(1-x/6)^5

peaceboy

酱以你的人生经验,
integrate 爱 dx = ???
kelfaru 发表于 25-9-2010 04:59 PM

    爱x+C

kelfaru

(3-x/2)^5 in ascending powers of 1/x
=(3/x-x^2)^5
hongji 发表于 25-9-2010 05:17 PM

第一个应该是酱...

(3 - x/2)^5
=[x(3/x - 1/2)]^5
=[(x/2)(6/x - 1)]^5
=[(-x/2)(1 - 6/x)]^5
=(-x/2)^5 (1 - 6/x)^5

Allmaths

爱x+C
peaceboy 发表于 25-9-2010 06:29 PM

   看来这位施主已体验过...

wuhu

看来这位施主已体验过...
Allmaths 发表于 25-9-2010 06:40 PM

   好怪啊～

hongji

1)find the coefficient of the term indicated in brackets in the expansion of each of the expanssions below
(a)  (x^3+1/x^2)^7.......(x^11)


(b)(2/x-x^4)^5.............(x^10)



2)x^2(4/x+x^3)^10.......(constant or independent of x)

3)expand the function √x^2+3   -  √x^2+2 in ascending powers of 1/x as far as he term in 1/x^3

4)find the constants A,B,C,D such that

      3x^2+5x/(1-x^2)(1+2)^2=A/1-x+B/1+x+C/(1+x)^2+D/(1+)^3

peaceboy

1)find the coefficient of the term indicated in brackets in the expansion of each of the expanssions ...
hongji 发表于 26-9-2010 05:42 PM

     (x^3+1/x^2)^7

T n+1=7Cn(x^3)^n (1/x^2)^(7-n)=7Cn x^3n (1/x^2)^7 (1/x^2)^(-n)
                                     =7Cn x^3n (x)^(-14) (x^2n)
                                     =7Cn x^(3n+2n-14)

3n+2n-14=11
5n=25
n=5

coefficient=7C5=


.................

(2/x-x^4)^5

Tn+1 = 5Cn (2/x)^n (-x^4)^(5-n)
       =5Cn (2)^n (x)^(-n) (-)^5 (x)^20 (-)^n (x)^(-4n)
       =5Cn (2)^n (x)^(-n+20-4n)(-)^(5+n)
      
-n+20-4n=10
5n=10
n=2

coefficient = (-)^(5+2) (5C2) (2)^2
               = -4 (5C2)         
.....

x^2(4/x+x^3)^10

Tn+1 = x^2 10Cn (4/x)^(n)  (x^3)^(10-n)
        =x^2  10Cn  4^n  x^(-n)  x^30  x^(-3n)
          =10Cn 4^n  x^(2-n+30-3n)

2-n+30-3n=0
4n=32
n=8

coefficient = 10C8 (4)^8

................

√x^2+3   -  √x^2+2 = (x^2+3)^(1/2) -(x^2+2 )^(1/2)
                              = x (1+3/x^2)^(1/2) - x (1+2/x^2)^(1/2)

然后bianomial expansion  (1+3/x^2)^(1/2)和(1+2/x^2)^(1/2)

..........

4问题有错么？

nkrealman

不错的TRIAL PAPER,试试看～
kelfaru 发表于 24-9-2010 10:54 AM

    可以给我完整的paper 1跟 paper2 math T吗？谢谢

peter_chu

sorry! 我有问题!
1)in the binomial expansion of (1+1/3 x)^n , the coefficients of the fourth and fifth term are equal. Find the value of n.
2)the coefficient of x^5 in binomial expansion of (1+5x)^8 id the same as the coefficient of x^4 in the expansion of (a+5x)^7.find the value of a.
3)if the first three terms in the expansion of (1+ax)^n in ascending powers of x are 1-4x+7x^2, find n and a.
4)by using the identity r^4-(r-1)^4 ≡ 4r^3-6r^2+4r-1 and ∑_(r=1)^n&#9618;〖r^6〗=1/6 n(n+1)(2n+1).find ∑_(r=1)^n&#9618;〖r^3〗

kelfaru

回复 1772# nkrealman

peaceboy

sorry! 我有问题!
1)in the binomial expansion of (1+1/3 x)^n , the coefficients of the fourth and fi ...
peter_chu 发表于 28-9-2010 12:18 AM

    1)in the binomial expansion of (1+1/3 x)^n , the coefficients of the fourth and fifth term are equal. Find the value of n.

T4=nC3 (1)^3 (x/3)^(n-3)
    =nC3 (1/3)^(n-3)  x^(n-3)

T5 =nC4 1^4 (x/3)^(n-4)
= nC4 (1/3)^(n-4) x^(n-4)

nC3 (1/3)^(n-3) =nC4 (1/3)^(n-4)

nC3 (1/3)^(n-3) / nC4 (1/3)^(n-4)  =1
3!(n-3)! / 4!(n-4)!  =3
3!(n-3)/4! =3
n-3 = 3*4!/3!
      =12
n=15

2)the coefficient of x^5 in binomial expansion of (1+5x)^8 id the sameas the coefficient of x^4 in the expansion of (a+5x)^7.find the valueof a.

8C3 (1)^3 (5x)^5 = 8C3 5^5 x^5
7C3 a^3 (5x)^4 = 7C3 a^3 5^4 x^4
8C3* 5^5 = 7C3 *a^3* 5^4

自己搬

3)if the first three terms in the expansion of (1+ax)^n in ascending powers of x are 1-4x+7x^2, find n and a.
1 +(n)/1! * (ax) + (n)(n-1)/2! *(ax)^2 =  1-4x+7x^2
by comparison
(n)/1! * (ax) =-4x ---equation 1
(n)(n-1)/2! *(ax)^2=7x^2 ---- equation 2

然后自己来

4)by using the identity r^4-(r-1)^4 ≡ 4r^3-6r^2+4r-1 and ∑_(r=1)^n&#9618;〖r^6〗=1/6 n(n+1)(2n+1).find ∑_(r=1)^n&#9618;〖r^3〗
&#9618;〈〈看不懂


最近在预考，比较懒惰

kelfaru

sorry! 我有问题!
1)in the expansion of (1+1/3 x)^n , the coefficients of the fourth and fi ...
peter_chu 发表于 28-9-2010 12:18 AM

1)Using Tr+1 = nCr a^(n-r) b^r

Term 4
T3+1 = nC3 (x/3)^3
Term 5
T4+1 = nC4 (x/3)^4

*coefficients of T5=T4

(nC3)/21 = (nC4)/81
324/27 = n-3
n=15   

2) Same concept
T5+1 = 8C5 (5x)^5
= 56(25x^2)
= 1400x^2

T4+1 = 7C4 a^(7-4) (-5x)^4
= 35a^3(625x^4)
= 21875a^3x^4

1400=21875a^3
8/125=a^3
a=2/5

3)
(1+ax)^n = 1 + n(ax) + n(n-1)/2! (ax)^2 + n(n-1)(n-2)/3! (ax)^3...

Given 1 - 4x + 7x^2 + ....

n(ax) = -4x
na = -4
a= -4/n------------*

n(n-1)/2! (ax)^2 = 7x^2
n(n-1)/2! a^2 = 7

Subt * inside
n(n-1)/2! (-4/n)^2 =7
n(n-1)/2 (16/n^2) = 7
8(n-1)/n = 7
8n - 8 = 7n
n=8

From *
a= -4/n
a= -1/2
第四题可以TYPE清楚一点吗???

kelfaru

回复 1775# peaceboy


妖,要做早讲一点...

hongji

(x^3+1/x^2)^7

T n+1=7Cn(x^3)^n (1/x^2)^(7-n)=7Cn x^3n (1/x^2)^7 (1/x^2)^(-n)
         ...
peaceboy 发表于 26-9-2010 10:51 PM

应该没有吧
不过我有两个超不多一样的问题
第一个：3x^2+5x/(1-x^2)(1+x^2)=A/1-x+B/1+x+c/(1+x)^2+D/(1+x)^3
第二个：.......         ........(1+x)^2   .........................

你拿5科全部学校都有教吗？

peaceboy

应该没有吧
不过我有两个超不多一样的问题
第一个：3x^2+5x/(1-x^2)(1+x^2)=A/1-x+B/1+x+c/(1+x)^2+D/( ...
hongji 发表于 28-9-2010 09:25 PM

第一的应该做不到
  2)  (3x^2+5x)/(1-x^2)(1+x)^2=A/(1-x)+B/(1+x)+c/(1+x)^2+D/(1+x)^3

(3x^2+5x)/(1-x^2)(1+x)^2= (3x^2+5x)/(1-x)(1+x)(1+x)^2
                                      = (3x^2+5x)/(1-x)(1+x)^3

A/(1-x)+B/(1+x)+c/(1+x)^2+D/(1+x)^3 =[A(1+x)^3 + B(1-x)(1+x)^2 + C (1-x)(1+x) + D(1-x)]/(1-x)(1+x)^3

(3x^2+5x)/(1-x)(1+x)^3=[A(1+x)^3 + B(1-x)(1+x)^2 + C (1-x)(1+x) + D(1-x)]/(1-x)(1+x)^3

by comparison ,
3x^2+5x=A(1+x)^3 + B(1-x)(1+x)^2 + C (1-x)(1+x) + D(1-x)

when x=1 , 8=8A
               A=1

when x=-1 , -2 = 2D
                   D=-1

when x=0 , 0=1+B+C-1
                   B+C = 0 ---1

when x = -2 , 2 = -1+3B-3C-3
                     3B-3C = 6
                       B-C=2 ----2

1+2 , 2B =2
           B=1

from 1 ,  B+C = 0
C=0-1
= -1

一科学校没教，自己外面补习

Allmaths

今天考的paper...