# 一些國中數學難題

1.[2^2/(1*3)]*[4^2/(3*5)]*[6^2/(5*7)]*......*[20^2/(19*21)]=?

2.前兩位數字與後兩位數字分別組成的二位數之和的平方，恰好等於這個四位數，問此四位數字為何？

3.設S=[2/(1*3*5)]+[2^2/(3*5*7)]+[2^3/(5*7*9)]+......+[2^48/(95*97*99)]

T=[1/(1*3)]+[2/(3*5)]+[2^2/(5*7)]+......+[2^48/(97*99)]

[2^2/(1*3)]*[4^2/(3*5)]*[6^2/(5*7)]*......*[20^2/(19*21)]要改成[2^2/(1*3)]+[4^2/(3*5)]+[6^2/(5*7)]+......+[20^2/(19*21)]=?

### 2 個解答

• 天助
Lv 7
1 0 年前
最佳解答

Q3的T是否有錯?

2010-04-24 22:36:40 補充：

Q1:

[2^2/(1*3)]+[4^2/(3*5)]+[6^2/(5*7)]+......+[20^2/(19*21)]

=[1+ 1/(1*3)]+[1+1/(3*5)]+...+[1+1/(19*21)]

=10+0.5[(1/1 -1/3)+(1/3-1/5)+...+(1/19- 1/21)]

=10+0.5[1/1- 1/21]= 10+ 10/21= 220/21

Q2:

設四位數2位數為a,末2位數為b,then原數=100a+b

題目: (a+b)^2=100a+b, so, (a+b-50)^2= -99b+2500 ---(A)

thus 2500-99b為完全平方(正數or 0), b <= 25

又 2500-99b為完全平方數,then b=01 or 25 代入(A)

b=01, (a-49)^2=2401=49^2, then a=0(不合) or 98 , 得四位數=9801

b=25, (a-25)^2=25, then a=20 or 30, 得四位數=2025 or 3025

Ans: 2025, 3025, 9801

Q3:

S=[2/(1*3*5)]+[2^2/(3*5*7)]+[2^3/(5*7*9)]+......+[2^48/(95*97*99)]

4S= [2/(1*3)- 2/(3*5)]+[2^2/(3*5)-2^2/(5*7)]+...+[2^48/(95*97)- 2^48/(97*99)]

=[2/(1*3)]+[2/(3*5)]+[2^2/(5*7)]+...+[2^47/(95*97)]- [2^48/(97*99)]

T=[1/(1*3)]+[2/(3*5)]+[2^2/(5*7)]+...+[2^47/(95*97)]+[2^48/(97*99)]

so, 4S-T=1/(1*3) - 2^49/(97*99)

則12S-3T=3(4S-T)=1- 2^49/(97*33)

Note:If T=1/(1*3)+2/(3*5)+...+2^47/(95*97) － 2^48/(97*99)

then 12S-3T=1

• 1 0 年前

沒有ㄟ..題目就是這樣給