Transforming Equation

[fonixbob]

Given that,

C(n+1) = (1+r)[C(n) - (1+i)^n * S]

How can I prove that:

C(n) = (1+r)^n [ Co - ((1+r)/(r-i))* S] + ((1+r)/(r-i))*(1+i)^n * S


I get really confused with the exponents

 

[Denis]

Given that,
C(n+1) = (1+r)[C(n) - (1+i)^n * S]
How can I prove that:
C(n) = (1+r)^n [Co - ((1+r)/(r-i))* S] + ((1+r)/(r-i))*(1+i)^n * S

Looks to me like that comes from some financial formula; C = Coupons?...do you know which?

Anyway, what is "Co" ?????

May be less confusing/unwieldy if you let a = 1+r, b = 1+i :
C(n+1) = a[C(n) - b^n * S]

C(n) = a^n [Co - a/(r-i) * S] + a/(r-i) * b^n * S

That's the extent of what I can do, as I'm not sure what you're doing :shock: