# Proof by induction! Urgh! :(

• Nov 20th 2007, 02:13 PM
Jimbobobo
Proof by induction! Urgh! :(
edit
11^2 + 12^1 = 133
11n+3 + 12^2n+3
But then I get stuck!? Any ideas?
• Nov 20th 2007, 02:42 PM
Plato
Say it is true for K>1, $\left( {11} \right)^{K + 2} + \left( {12} \right)^{2K + 1} = 133j$.
Look at the case of K+1.
$\left( {11} \right)^{\left( {K + 1} \right) + 2} + \left( {12} \right)^{2\left( {K + 1} \right) + 1}$
$= \left( {11} \right)^{K + 3} + 11\left( {12} \right)^{2K + 1} - 11\left( {12} \right)^{2K + 1} + \left( {12} \right)^{2K + 3}$
$\begin{array}{l}
= 11\left[ {\left( {11} \right)^{K + 2} + \left( {12} \right)^{2K + 1} } \right] + \left( {12} \right)^{2K + 1} \left[ { - 11 + \left( {12} \right)^2 } \right] \\
= 11\left[ {133j} \right] + \left( {12} \right)^{2K + 1} \left[ {133} \right] \\
\end{array}
$
• Nov 21st 2007, 08:30 AM
topsquark
Plato either wins the award for most psychic Math Helper, or Jimbobobo edited his original question to the point where it is meaningless. I'm curious as to which?

-Dan
• Nov 21st 2007, 08:43 AM
Plato
The original question was edited.