Simplify formula by removing sum from it I have a question about the following formula, I just need to write it without a sum with some formula convertion(simplify it). As for now I'm stuck, I don't know if it is possible for now, but hope it is.
$$
\sum_{i=1}^l \frac{b_i^2}{1+AB_i}, \quad \text{where }B_i=\sum_{\textstyle j= \, i}^L b_{\textstyle j}, \  \text{and } \ l\leq L
$$
I tried to lead to a common denominator, make some assumptions to simplify the formula, but for now I'm stuck, any help is appreciated.
$$
\sum_{i=1}^l \frac{b_i^2}{1+AB_i}= \frac{b_1^2}{1+A(b_1+...+b_L)} + \frac{b_2^2}{1+A(b_2+...+b_L)}+...+\frac{b_l^2}{1+A(b_l+...+b_L)}
$$
 A: There are $2D$ sums in the expression, which can be transformed, using formulas for the negative binomials
$$\frac1{(1-x)^{m+1}} = \sum_{n=m}^\infty \binom nm x^{n-m}$$
which allow to write according polynomial formulas, wherein the polynomial coefficients equal to the binomial coefficients production.
Simplification of the polynomial formulas requires additional assumptions about given coefficients (positive? sorted?$\dots$ Are
$Ab_i < 1?$ etc.)
Partially, if $b_i\in(0,1)$ and $0<A<<1,$ then the approximation
$$1+A\sum b_i\approx \prod(1+Ab_i)$$
allows to obtain suitable common denominators.
Another essential advantages has the approximation 
$$1+A\sum b_i \approx e^{A b_i}.$$
If the sequence $\{b_l\}$ increases and $0<A<<1$ or $L>>l,$ i.e. if $1+AB_l > A(B_1-B_l),$ then looks correct the next approach.
Denote $D_i=1+AB_i,\quad D = D_l,\quad \alpha\,=\dfrac A{D_l},$ then
\begin{align}
&D_{l-1}= 1+AB_{l}+Ab_{l-1} = (1+AB_l)\left(1+\dfrac{A}{1+AB_l}b_{l-1}\right) \approx D e^{\alpha\, b_{l-1}},\\[4pt]
&D_{l-2}\approx D_{l-1} e^{\alpha\,b_{l-2}} = De^{\alpha\,(B_{\,l-2}-B_{\,l})},\dots,\\[4pt]
&D_{i}\approx D_{i+1} e^{\alpha\,b_i} = De^{\alpha\,(B_{\,i}-B_{\,l})},\dots,\\[4pt]
&D_{1}\approx D_{2} e^{\alpha\,b_1} = De^{\alpha\,(B_{\,1}-B_{\,l})},\\[4pt]
&\sum_{i=1}^l\dfrac{b_i^2}{1+AB_i} 
\approx \sum_{i=1}^l\dfrac{b_i^2}{D e^{\alpha\,(B_{\,i}-B_{\,l})}}, 
\end{align}
$$S(\alpha) = \sum_{i=1}^l\dfrac{b_i^2}{1+AB_i} \approx \dfrac1{D}\sum_{i=1}^l b_i^2e^{-\alpha S_i},\quad\text{where}\quad S_i=\sum_{j=i}^{l}b_j.$$
Obtained expression looks more suitable for the further detalizations. But I cannot estimate real progress.
On the other hand,
\begin{align}
&D_{l-1}= 1+AB_{l}+Ab_{l-1} = (1+AB_l)\left(1+\dfrac{A}{1+AB_l}b_{l-1}\right),\\[4pt]
&D_{l-2}= 1+AB_{l}+Ab_{l-1}+Ab_{l-1} = D\left(1+\alpha b_{l-2}+\alpha b_{l-1}\right)\approx D\left(1+\alpha b_{l-2}\right)\left(1+\alpha b_{l-1}\right),\dots,\\[4pt]
&D_{i}\approx D\left(1+\alpha b_{i}\right)\left(1+\alpha b_{i+1}\right)\dots\left(1+\alpha b_{l-1}\right),\dots,\\[4pt]
&D_{1}\approx D\left(1+\alpha b_{1}\right)\left(1+\alpha b_{2}\right)\dots\left(1+\alpha b_{l-1}\right),\\[4pt]
\end{align}
$$S(\alpha) \approx \dfrac1{D}\sum_{i=1}^l b_i^2T_i,\quad\text{where}\quad T_l=\dfrac1D,\quad T_i = \dfrac {T_{i+1}}{1+\alpha b_i}.$$
