Can you solve this geometric progression to infinity?

[xzuberent]

I'm trying to solve the following series to be a single equation which I think is

V = Σ [C0(1+g)n+3]/(1+r)n).

Can you solve the below geometric progression series with the 'g' component increasing every series of 3? Please show your step by step working method:

V = C0/(1+r) + C0 /(1+r)2 + C0 /(1+r)3 + C0(1+g)3/(1+r)4 + C0(1+g)3/(1+r)5 +
C0(1+g)3/(1+r)6 + C0(1+g)6/(1+r)7 + ……

Thanks!

[shyamjayakannan]

you can divide this into 3 separate series:
1) [tex]\frac{C_0}{1+r}+\frac{C_0(1+g)^3}{(1+r)^4}+\frac{C_0(1+g)^6}{(1+r)^7}...=\frac{C_0}{1+r}\sum_{i=0}^\infty\left(\frac{1+g}{1+r}\right)^{3i}[/tex]

2) [tex]\frac{C_0}{(1+r)^2}+\frac{C_0(1+g)^3}{(1+r)^5}+\frac{C_0(1+g)^6}{(1+r)^8}...=\frac{C_0}{(1+r)^2}\sum_{i=0}^\infty\left(\frac{1+g}{1+r}\right)^{3i}[/tex]

3) [tex]\frac{C_0}{(1+r)^3}+\frac{C_0(1+g)^3}{(1+r)^6}+\frac{C_0(1+g)^6}{(1+r)^9}...=\frac{C_0}{(1+r)^3}\sum_{i=0}^\infty\left(\frac{1+g}{1+r}\right)^{3i}[/tex]

Adding all three, we get: [tex]\left\{\frac{C_0}{1+r}+\frac{C_0}{(1+r)^2}+\frac{C_0}{(1+r)^3}\right\}\sum_{i=0}^\infty\left(\frac{1+g}{1+r}\right)^{3i}=\boxed{\frac{C_0\left(r^2+3r+3\right)}{(1+r)^3}\sum_{i=0}^\infty\left(\frac{1+g}{1+r}\right)^{3i}}[/tex]