2條F5 Math Question-－ohe09nz81r的部落格

標題:

2條F5 Math Question*

發問:

1. Find the sum of the first 3 terms of a geometric series 4+6+9+...that can give a sum greater than 800. 2. How many terms of each of the following geometric series must be taken to give the sum stated? 36-18+9-...=23又128分之125 THZ~~~^^

最佳解答:

1.) common ratio=1.5 let a be the first term a + 1.5a + 2.25a > 800 4.75a > 800 a > 168.42 if all 3 terms must be integers, then a = 172 so, the sum of the first 3 terms of a geometric series is 172+258+387=817 2.) 36-18+9-...=23又128分之125 therefore, first term = a = 36 and common ratio = R = -0.5 let n be the number of terms [a(1-R)^n]/(1-R) = 3069/128 [36(1-(-0.5)^n)]/(1+0.5) = 3069/128 24[1-(-0.5)^n] = 3069/128 1-(-0.5)^n = 1023/1024 1-1023/1024 = (-0.5)^n 1/1024 = (-0.5)^n (0.5)^10 = (-0.5)^n n = 10 so, the first 10 terms of the geometric series = 23又128分之125 備註: (-0.5)^n = (-0.5)的n次 1023/1024 = 1024分之1023 2007-10-08 17:25:49 補充： 1.) this question is talking about geometric series, not Arithmetic Series~~ 2007-10-08 19:20:29 補充： 2.) 其實可以check下答案~~first 10 terms: 36, -18, 9, -9/2, 9/4, -9/8, 9/16, -9/32, 9/64, -9/12836 - 18 9 - 9/2 9/4 - 9/8 9/16 - 9/32 9/64 - 9/128= 36 - 18 9 - 4.5 2.25 - 1.125 0.5625 - 0.28125 0.140625 - 0.0703125= 23.976562523又128分之125 = 23.9765625so, 一定係10個terms~

其他解答:

n th the sum of a geometric series = 1/2n[2a+(n-1)d] 1.a=4 d=2 so 1/2n[2(4)+(n-1)(2)] greader than 800 n^2 + 3n - 800 greader than 0 n greader than 26.824 or -29.824 greader than n(rej) n=27 n th the sum of等級(呢個唔識eng)=[a(r^n-1)]/(r-1) 2. a=36 r=-1/2 because 36 * -1/2 =-18 so 23又128分之125=[(36)((-1/2)^n-1)]/((-1/2)-1) 23又128分之125=[(36)((-1/2)^n-1)]/(-3/2) 23又128分之125=(-24)[(-1/2)^n-1] 23又128分之125=(-24)(-1/2)^n+24 (-24)(-1/2)^n=-128分之1 (-1/2)^n=3072分之1 log(-1/2)^n=log3072分之1 (n+1)．log(1/2)=log3072分之1 n=10.5849625 n=11