little problem with CAS

[Thomas Klemm]

(01-25-2015 08:52 PM)Hlib Wrote:  \((6+1)*(6^2+1)*(6^4+1)*(6^8+1)*(6^{16}+1)-0.2*6^{32} =?\)

You can multiply the left product by \(a-1\) which consecutively "eats up" the next factor:
\[
\begin{align}
(a-1)(a+1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1) & \\
(a^2-1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1) & \\
(a^4-1)(a^4+1)(a^8+1)(a^{16}+1) & \\
(a^8-1)(a^8+1)(a^{16}+1) & \\
(a^{16}-1)(a^{16}+1) & \\
(a^{32}-1) & \\
\end{align}
\]

Thus we end up with:
\[\frac{a^{32}-1}{a-1}-\frac{a^{32}}{a-1}=\frac{-1}{a-1}\]

This is \(\frac{-1}{5}\) for \(a=6\).

Cheers
Thomas