little problem with CAS

[Hlib]

(01-30-2015 08:40 AM)Angus Wrote:  ... but I can't figure out how 0.2*a^32 converts to a^32/(a-1) ?

Thank you

\[\frac{a^{32}-1}{a-1}-\frac{a^{32}*(a-1)}{5*{(a-1)}}\]
\[\frac{a^{32}-1}{a-1}-\frac{a^{32}*(6-1)}{5*{(a-1)}}=\frac{-1}{a-1}\]
\[\frac{a^{32}-1}{a-1}-\frac{a^{32}}{a-1}=\frac{-1}{a-1}\]
Thomas Klemm forgot to write "5". This is a groundless mathematical trick, but the result is really true.