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I have an expression that is similar to this:

$$L_0 + (L_0 + L_1) + (L_0 + L_1 + L_2) + (L_0 + L_1 + L_2 + L_3) + .... + (L_0 + L_1 + ... + L_n)$$

How would I be able to simplify this recursive expression?

suomynonA
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Naz
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  • This is the sum of the partial sums up to $n$ of $(L_n)$. The most you could do to "simplify" it would be $$(n+1)L_0 + nL_1 + \cdots + 2L_{n-1} + L_{n}$$ –  Oct 26 '16 at 03:11
  • You could rewrite it in sigma notation $ \sum_{i=0}^n \sum_{j=0}^i L_j $ – Q the Platypus Oct 26 '16 at 03:30
  • Based on Bye_World's expression the expression could be placed into a summation form as: $\sum_{k=0}^{n} (n-k+1) , L_{k}$ or $\sum_{k=0}^{n} (k+1) , L_{n-k}$ – Leucippus Oct 26 '16 at 04:26

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As said in comments, you may consider that the expression write $$S_n=\sum_{k=0}^n s_k\qquad \text{using}\qquad s_k=\sum _{i=0}^k L_i$$ which makes $$S_n=\sum _{k=0}^n \left(\sum _{i=0}^k L_i\right)$$ If the $L_i$'s are in arithmetic progression $(L_i=a+id)$ this will make $$s_k=\frac{1}{2} (k+1) (2 a+d k)\implies S_n=\frac{1}{6} (n+1) (n+2) (3 a+d n)$$ If the $L_i$'s are in geometric progression $(L_i=ar^{i})$ this will make $$s_k=a\frac{ \left(r^{k+1}-1\right)}{r-1}\implies S_n=a\frac{ r^{n+2}-(n+2) r+(n+1)}{(r-1)^2}$$ If the $L_i$'s are in harmonic progression $(L_i=\frac{1}{a+id })$ this will make $\cdots$ a nightmare.

Claude Leibovici
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