數學高手請進!!!

a)
On the first day of the month you would get 1c = 2(1-1)c,
On the second day you would get 2c = 2(2-1)c,
On the third day 4c = 2(3-1)c,
On the fourth day 8c = 2(4-1)c,
So on the nth day, you would get 2(n-1)c
And on the 30th day, you would get 2(30-1)c = 229c

b)
Amount received in total
= c + 2c + 22c + 23c + … + 229c
= c (1 + 2 + 22 + 23 + … + 229 )
= c (230 – 1) / (2 – 1)
= (230 – 1) c

2007-09-08 19:31:15 補充：
Since (x – 1) [ x^(n-1) x^(n-2) … x 1 ]= x * x^(n-1) x * x^(n-2) … x * x x * 1 - [ x^(n-1) x^(n-2) … x 1 ]= x^n x^(n-1) … x^2 x –x^(n-1) – x^(n-2) – … – x – 1= x^n – 1So [ x^(n-1) x^(n-2) … x 1 ] = (x^n – 1) / (x – 1)

2007-09-08 19:31:34 補充：
By substituting x = 2 and n = 30, we have (1 2 2^2 2^3 … 2^29 ) = (2^30 – 1) / (2 – 1) = 2^30 – 1

2007-09-08 21:51:44 補充：
唔見吱符號，再來一次：Since (x – 1) [ x^(n-1) + x^(n-2) + … + x + 1 ]= x * x^(n-1) + x * x^(n-2) + … + x * x + x * 1 - [ x^(n-1) + x^(n-2) + … + x + 1 ]= x^n + x^(n-1) + … + x^2 + x –x^(n-1) – x^(n-2) – … – x – 1= x^n – 1So [ x^(n-1) + x^(n-2) + … +x + 1 ] = (x^n – 1) / (x – 1)

2007-09-08 21:52:44 補充：
By substituting x = 2 and n = 30, we have (1 + 2 + 2^2 + 2^3 + … + 2^29 ) = (2^30 – 1) / (2 – 1) = 2^30 – 1