每年存一元 存n年, 年息x%
(1+x）^（n-1）+（1+x）^（n-2）+...+(1+x）^0
=>
1） Sn = S(n-1) + (1+x）^(n-1)
2）(1+x）* S(n-1）= Sn - (1+x)
1) * (1+x) then substituted by 2) =>
(1+x)*Sn = (1+x)*S(n-1) + (1+x)^n
=>
(1+x)*Sn = Sn - (1+x) + (1+x)^n
=>
Sn = [(1+x)^n - (1+x)]/x
If n = 40, x = 7%
S40 = 198.63
110/m = 1320/year
Money we can have after 40 years:
1320 * 198.63 = 262192, which is less than 300000!
(1+x）^（n-1）+（1+x）^（n-2）+...+(1+x）^0
=>
1） Sn = S(n-1) + (1+x）^(n-1)
2）(1+x）* S(n-1）= Sn - (1+x)
1) * (1+x) then substituted by 2) =>
(1+x)*Sn = (1+x)*S(n-1) + (1+x)^n
=>
(1+x)*Sn = Sn - (1+x) + (1+x)^n
=>
Sn = [(1+x)^n - (1+x)]/x
If n = 40, x = 7%
S40 = 198.63
110/m = 1320/year
Money we can have after 40 years:
1320 * 198.63 = 262192, which is less than 300000!