等比數列之和...help

設數列為T ( n ) = arn-1,
T ( n ) = 16, 所以
a( 2 / 3 )n-1 = 16
a = 16 / ( 2 / 3 )n-1 --- ( 2 )
S ( n ) = 211, 所以
a[ 1 – ( 2 / 3 )n ] / ( 1 – 2 / 3 ) = 211 --- ( 2 )
代( 1 )入( 2 ),
[16 / ( 2 / 3 )n-1] [ 1 – ( 2 / 3 )n ] / ( 1 – 2 / 3 ) = 211
[16[ 1 – ( 2 / 3 )n ] / ( 2 / 3 )n-1] = 211 / 3
[ 1 – ( 2 / 3 )n ] / ( 2 / 3 )n-1 = 211 / 48
211 ( 2 / 3 )n-1 + 48 ( 2 / 3 )n = 48
( 2 / 3 )n [ 211 ( 2 / 3 )-1 + 48 ] = 48
( 2 / 3 )n = 32 / 243
( 2 / 3 )n = ( 2 / 3 )5
n = 5
a = 16 / ( 2 / 3 )5-1 = 81
所以該數列的首項是81, n是5。

a1*r^n-1=16
[a1*(1-r^n)]/(1-r)
= a1-a1*r^n/(1-2/3)
= 211
a1*r^n= (a1*r^n-1)*n=16*(2/3)=32/3
a1-(32/3)/(1/3)=211
a1=211/3 + 32/3=81

81*(r^n-1)=16
(2/3)^n-1=16/81
n-1=4 n=5

a)a1=81
b)n=5