little problem with CAS
01-25-2015, 10:17 PM
Post: #4
 Thomas Klemm Senior Member Posts: 2,154 Joined: Dec 2013
RE: little problem with CAS
(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
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 Messages In This Thread little problem with CAS - Hlib - 01-25-2015, 08:52 PM RE: little problem with CAS - Mark Hardman - 01-25-2015, 09:04 PM RE: little problem with CAS - Hlib - 01-25-2015, 09:53 PM RE: little problem with CAS - Thomas Klemm - 01-25-2015 10:17 PM RE: little problem with CAS - Hlib - 01-26-2015, 08:22 AM RE: little problem with CAS - Katie Wasserman - 01-25-2015, 10:51 PM RE: little problem with CAS - rprosperi - 01-26-2015, 01:23 AM RE: little problem with CAS - Angus - 01-30-2015, 08:40 AM RE: little problem with CAS - Hlib - 01-30-2015, 01:00 PM

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