(a)Find the sum of the following series in terms of n:
1+3+3^2+3^3+.......+3^n
(b)Find the sum of the following series in terms of n:
1+1/3+1/3^2+1/3^3+.......+1/3^n
(c)Find the sum of the first 11 terms of the series:
1+3/1/3+9/1/9+27/1/27+correct to 1decimal place
3/1/3=3and1/3
answers:
(a)1/2[(3^n+1)-1]
(b)3/2[1-(1/3)^n+1]
(c)88573.5
How to get it??pls.
sum of series?
2007-06-12 5:50 pm
回答 (2)
2007-06-12 6:26 pm
✔ 最佳答案
sum of G.S.=a*(r^N-1)/(r-1)OR
sum of G.S.=a*(1-r^N)/(1-r)
a=first term
N=no.of terms
r=ratio
(a)
a=1
N=n+1
r=3
1+3+3^2+3^3+.......+3^n
=1*[3^(n+1)-1]/(3-1)
=1/2*[3^(n+1)-1]
(b)
a=1
N=n+1
r=1/3
1+1/3+1/3^2+1/3^3+.......+1/3^n
=1*[1-(1/3)^(n+1)]/(1-1/3)
=[1-(1/3)^(n+1)]/(2/3)
=3/2*[1-(1/3)^(n+1)]
(c)
sum of the first 11 terms of the series 1+3/1/3+9/1/9+27/1/27+...
=sum of the first 11 terms of the series
1+3+1/3+3^2+1/3^2+3^3+1/3^3+...
=sum of the first 11 terms of the series
1+3+3^2+3^3+.......+1/3+1/3^2+1/3^3+.......
=1/2*[3^(11)-1]+3/2*[1-(1/3)^(11)]-1
(using the result of (a) & (b))
=88.573.5
(correct to 1decimal place)
參考: me
2007-06-12 6:17 pm
參考: My Maths knowledge
收錄日期: 2021-04-12 19:54:32
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https://hk.answers.yahoo.com/question/index?qid=20070612000051KK00672