# 數學geometric pro-gression緊急!!!

Question 2

An arithmetic series has first term a and common difference d. The sum of the first 19 terms is 266.

(a) Show that a + 9d = 14.

(b) The sum of the fifth and eighth terms is 7. Find the values of a and d.

----------------------------------------

Question 3

The cost of a life insurance policy is £55.89 in the first year. Subsequently the cost in each year is 3% more than the cost in the previous year.

(a) Show that the yearly costs of this life insurance policy form a geometric pro-gression.

(b) Find the cost of the policy in the twentieth year.

(c) Find the total of all the payments made by the end of the twentieth year

Update:

(a) Whrite down the binomial expansion of (2+x)的5次方. Hence, express (2-耕號3)的5次方 in the form of p+q耕號3, where p and q are integers. State the values of p and q.

(b) Find the coefficient of the x的7次方 term in the binomial expansion of (x的3次方+3x)的5次方

Update 2:

Rating

Question 2

(a) The sum of the first 19 terms is 19*(a+a19)/2=19*[a+(a+18d)]/2=19*(2a+18d)/2=266 (divide by 19 on both sides)

=> (2a+18d)/2 = 14

=> a+9d = 14

(b) the fifth term = a+4d, the eighth term = a+7d

(a+4d)+(a+7d) = 2a+11d=7

but we also know that 2a+18d=28 by (a)

Thus we have to equations to solve this problem:

2a+11d=7 ... eq1

2a+18d=28 ... eq2

eq2-eq1 => 7d=21 => d=3

2a+11*33=7 => 2a=7-33=-26 => a=-13

Question 3

(a)

the first year => 55.89

the second year => 55.89+55.89*0.03=55.89(1+0.03)=55.89*1.03

the third year => (55.89*1.03)+(55.89*1.03)*0.03=55.89*(1.03)^2

(keep going...)

the nth year => 55.89*(1.03)^(n-1), which forms a geometric progression.

(b) By the formula we just obtained from (a), the cost of the policy in the twentieth year is 55.89*(1.03)^19 = 98.0034533

(c) The formula of a geometric series is a(r^n-1)/(r-1).

Thus, the total of all the payments made by the end of the twentieth year is 55.89(1.03^20-1)/(1.03-1)=1501.785230.

2009-04-21 20:58:11 補充：

不曉得補充寫不寫得下欸...

C(n,m) = C n 取 m 這個你應該看得懂吧？先告訴你一下表示法。

(a)

Binomial expansion: (2+x)^5=C(5,0)*2^5+C(5,1)*2^4*x+C(5,2)*2^3*x^2+C(5,3)*2^2*x^3+C(5,4)*2*x^4+C(5,5)*x^5

2009-04-21 20:58:17 補充：

Let x=-(3^(1/2)) (這就是-根號3指數的表示法)

(2-(3^(1/2)))^5=1*32+5*16*(-(3^(1/2)))+10*8*3+10*4*(-3*(3^(1/2)))+5*2*9+1*(-9*(3^(1/2)))=32+240+90-80*(3^(1/2))-120*(3^(1/2))-9*(3^(1/2))=362-209*(3^(1/2))

所以p=362，q=-209

2009-04-21 20:58:20 補充：

(b)

(x^3+3x)^5=[x(x^2+3)]^5=x^5*(x^2+3)^5

外面已經有一個x的5次方了，要湊出x的7次方的話，(x^2+3)^5只需要x^2就可以了。

x^2項的係數就是取1個x^2配上4個3 =>係數= C(5,1)*3^4=5*81=405

因此x^7項的係數為405。