Consider this expression as rendered by default StandardForm in MMA (my installation,, Windows)

expr = ((3*2^Sum[Subscript[q, j], {j, 1, -1 + P}] + 3^P)*(2^Sum[Subscript[q, j], {j, 1, P}] + 3^P) + 3^(1 + 2*P)*n + 
  3*(2^Sum[Subscript[q, j], {j, 1, P}] + 3^P)*Sum[2^Sum[Subscript[q, j], {j, 1, i}]*3^(-i + P), {i, 2, -1 + P}])/
 (3*2^Sum[Subscript[q, j], {j, 1, P}])

When written out (here via TeXForm, which is not quite the same as StandardForm) it is readily apparent that the factor of $1/3$ cancels numerous explicit factors of $3$ and can be absorbed in factors of $3^m$ by the replacement $m \rightarrow m-1$

$\frac{1}{3} 2^{-\sum _{j=1}^P q_j} \left(3 \left(2^{\sum _{j=1}^P q_j}+3^P\right) \left(\sum _{i=2}^{P-1} 3^{P-i} 2^{\sum _{j=1}^i q_j}\right)+\left(3\ 2^{\sum _{j=1}^{P-1} q_j}+3^P\right) \left(2^{\sum _{j=1}^P q_j}+3^P\right)+n 3^{2 P+1}\right)$

The question is, how can I force MMA to make this simplification? Simplify does not do it, FullSimplify times-out.

If I use Expand[expr] I obtain

$3^P 2^{-\sum _{j=1}^P q_j} \sum _{i=2}^{P-1} 3^{P-i} 2^{\sum _{j=1}^i q_j}+\sum _{i=2}^{P-1} 3^{P-i} 2^{\sum _{j=1}^i q_j}+n 3^{2 P} 2^{-\sum _{j=1}^P q_j}+3^P 2^{\sum _{j=1}^{P-1} q_j-\sum _{j=1}^P q_j}+3^{2 P-1} 2^{-\sum _{j=1}^P q_j}+2^{\sum _{j=1}^{P-1} q_j}+3^{P-1}$

and then have factors such as $3^P$ that can/should be factored out again (where they occur in isolation; I would strongly prefer not to have $3^{p-1} \rightarrow 3^P /3$, for example} (Is there a pattern I could supply to Expand to avoid this?)

I suspect this could be done with replacement rules, but I fear the required rules would be rather ad hoc and would much prefer a (simpler) generic solution.

  • $\begingroup$ Related to getting 3^p out of the sum: (16969), (58091), (226003) -- Another point: 3^p/3 won't stay in that form. $\endgroup$
    – Michael E2
    Oct 18, 2022 at 11:24
  • $\begingroup$ @MichaelE2 Indeed! $3^P/3 \rightarrow 3^{-1+P}$, with leading -1, yet another - but separate - peeve! $\endgroup$ Oct 18, 2022 at 17:11


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