F fonixbob New member Joined Sep 6, 2010 Messages 3 Sep 7, 2010 #1 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
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
D Deleted member 4993 Guest Sep 9, 2010 #2 fonixbob said: 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 Click to expand... Are you writing a recursive relation like: \(\displaystyle C_{n+1} \ = \ (1+r)[C_n - (1+i)^n *S]\) please define r, i and S.
fonixbob said: 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 Click to expand... Are you writing a recursive relation like: \(\displaystyle C_{n+1} \ = \ (1+r)[C_n - (1+i)^n *S]\) please define r, i and S.
