math question

The sum of the first n terms
= 3[2^n - 1] / (2 - 1)...............[common ratio = 6 / 3 = 2.]
= 3[2^n - 1]
Therefore 3[2^n - 1] > 15000
2^n - 1 > 5000
2^n > 5001
log2^n > log5001
n log2 > log5001
n > log5001 / log2 = 12.288....
Thus the least value of n = 13.

the first term = a = 3
the ratio = r = 6/3 = 2
then, the n term, Tn = a*r^(n-1)

Sum of the first n terms of geometric sequence greater than 15000
a*(r^n - 1) / (r-1) greater than 15000
3*(2^n - 1) / (2-1) greater than 15000
3*(2^n - 1) greater than 15000
(2^n - 1) greater than 5000
2^n greater than 5001
n greater than (log 5001 / log 2)
n greater than 12.3
So than the least value of n is 13

a = 3 , r = 6/3 = 2

S(n) = a( r^n -1) / r-1 >15000
= 3( 2^n -1) / 2-1 >15000
= 3( 2^n -1) >15000
= 2^n -1 > 5000
= 2^n > 4999
because 2^13 > 4999 , n is 13

2006-10-31 23:35:52 補充：
= 2^n -1 = 5000= 2^n = 5001because 2^13 > 4999 , n is 13

3+6+12+24..... greater than 15000
3(2^n-1)/(2-1) greater than 15000
3x2^n-3 greater than 15000
2^n greater than 5001
n greater than 12.3,cor. to 3 sig. fig.

so the least value of n=13