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How to simplify the below balance equations? $$S_0 \left(\lambda _c+(1-2 q) \lambda _m\right)=\sum _{i=2}^{k-1} S_i \left(\lambda _c+(1-2 q) \lambda _m\right)+S_k \left(\lambda _c+(1-q) \lambda _m\right)+S_1 \left(\lambda _c+(1-q) \lambda _m\right)-2 \text{q$\lambda $}_m S_0 $$ $$S_1 \text{q$\lambda $}_m=2 S_0 \text{q$\lambda $}_m+S_2 \text{q$\lambda $}_m$$ $$S_2 \text{q$\lambda $}_m=S_1 \text{q$\lambda $}_m+S_3 \text{q$\lambda $}_m$$ $$S_3 \text{q$\lambda $}_m=S_2 \text{q$\lambda $}_m+S_4 \text{q$\lambda $}_m$$ $$\vdots$$ $$S_{k-1} \text{q$\lambda $}_m=S_{k-2} \text{q$\lambda $}_m+S_k \text{q$\lambda $}_m$$ $$S_{k-1} \text{q$\lambda $}_m=S_k \left(\lambda _c+(1-q) \lambda _m\right)+S_k \text{q$\lambda $}_m$$

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$S_{k-1} \text{q$\lambda $}_m=S_{k-2} \text{q$\lambda $}_m+S_k \text{q$\lambda $}_m$ -- is it the same as $S_{k-1} =S_{k-2} +S_k $ ? –  Boris Novikov Aug 2 '13 at 20:32

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