I am a little unsure what you are asking.
In your first sentence you say "each term is R percent larger then previous element", which leads to the recurrence $a_n = (1+R)a_{n-1}$ as DonAntonio and Arturo have analyzed.
But in your example each term is actually a constant factor larger than the sum of the previous two elements, which is $a_n = (1+R)(a_{n-1} + a_{n-2})$. This is a more difficult problem.
For convenience write $a_1 = \alpha$, $a_2 = \beta$, and $(1+R) = r$. Then the two-term recurrence gives \begin{align*} a_1 &= \alpha \\ a_2 &= \beta \\ a_3 &= r(a_1 + a_2) = r\alpha + r\beta \\ a_4 &= r(a_2 + a_3) = r^2\alpha + (r + r^2)\beta \\ a_5 &= r(a_3 + a_4) = (r^2 + r^3)\alpha + (2r^2 + r^3)\beta \\ a_6 &= r(a_4 + a_5) = (2r^3 + r^4)\alpha + (r^2 + 3r^3 + r^4)\beta \end{align*} and so on. The corresponding sums $S_N = \sum_{i=1}^N a_i$ are \begin{align*} S_1 &=\alpha \\ S_2 &=\alpha + \beta \\ S_3 &=(1+r)\alpha + (1+r)\beta \\ S_4 &=(1+r+r^2)\alpha + (1 + 2r + r^2)\beta \\ S_5 &=(1+r+2r^2+r^3)\alpha + (1+2r+3r^2+r^3)\beta \\ S_6 &=(1+r+2r^2+3r^3+r^4)\alpha + (1+2r+4r^2+4r^3 + r^4)\beta \end{align*} It is possible that we can spot a pattern here. Perhaps the coefficients in the expression before $\alpha$ are simply $1,1, 2, 3, ..., N-3, 1$ (though I don't see a nice pattern in the $\beta$ expression). Then, if we are lucky, we could find a closed form expression for $S_N$, which we could use to choose $\alpha$ and $\beta$ to give us a sum of 1.
But I am not optimistic.
Edit My lack of optimism was misplaced. If we take $\alpha = 2$ and $\beta = r$ (and index starting at zero rather than one) we have a (non-integer) Lucas sequence $V_n(r,-r)$: \begin{align*} V_0(r, -r) &= 2, \\ V_1(r, -r) &= r,\\ V_n(r, -r) &= rV_{n-1}(r,-r) + rV_{n-2}(r,-r). \end{align*} The analysis in the Wikipedia page shows that $V_n = a^n + b^n$ where $a = \frac{r + \sqrt{\Delta}}{2}, \quad b = \frac{r - \sqrt{\Delta}}{2}, \quad \Delta=r^2 + 4r$ and so the sum $S_N$ is \begin{align*} S_N &= \sum_{n=0}^N a^n + b^n \\ &= \frac{a^{N+1}-1}{a-1} + \frac{b^{N+1} -1}{b-1}. \end{align*} Then, by starting our sequence with $2/S_N$ and $r/S_N$ rather than with $2$ and $r$ we will get a sequence with a sum of 1.
Note, though, that this method doesn't allow us to choose $a_1/a_0$: the ratio must be $r/2$.