sum of series?

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)

As follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Crazyseries1.jpg