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# 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.

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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

還有Q.4

(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

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謝啦~ 你是我的普拿疼~

### 1 Answer

- TonyLv 51 decade agoFavorite Answer
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。