Help with recurrence relation It 's been a long time since I touched this kind of math , so please help me to solve the relation step by steps : 
$V_k = (1+i)*V_{k-1}+P$ 
I know the answer is  $V_k = (P/i)*((1+i)^k-1) $ 
Thanks in advance.

If you want a methodical way to find the answer that you don't know yet, you could use generating functions. Defining $F(x) = \sum_{k=0}^\infty  V_k x^k$, and assuming the initial condition is $V_0=0$, the recursion gives
\begin{align*}
F(x) &= V_0 x^0 + \sum_{k=1}^\infty \big( (1+i)V_{k-1} + P \big) x^k  \\
&= 0 + (1+i)x \sum_{k=1}^\infty V_{k-1} x^{k-1} + Px \sum_{k=1}^\infty x^{k-1} \\
&= (1+i)x \sum_{k=0}^\infty V_k x^k + Px \sum_{k=0}^\infty x^k \\
&= (1+i)x F(x) + \frac{Px}{1-x}.
\end{align*}
Solving for $F(x)$ gives $\big(1 - (1+i)x \big) F(x) = Px/(1-x)$, and so
\begin{align*}
F(x) &= \frac{Px}{(1 - (1+i)x )(1-x)} \\
&= \frac Pi \bigg( \frac1{(1-(1+i) x)}-\frac1{1-x} \bigg) \\
&= \frac Pi \bigg( \sum_{k=0}^\infty ((1+i) x)^k - \sum_{k=0}^\infty x^k \bigg).
\end{align*}
Comparing coefficients of $x^k$ then yields $V_k = \frac Pi \big( (1+i)^k - 1 \big)$.
