Challenging Mathematical Induction Question

[Vertciel]

Hello there,

I am having trouble with the following question on mathematical induction. My work is shown below.

Thank you very much.

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1. Prove that $\displaystyle E(n) = 11^{n + 2} + 12^{2n + 1}$ is divisible by 133 for all positive integers n.

Show $\displaystyle t_1$ :

E(1) = 1331 + 1728 = 3059. This is true because 3059 is divisible by 133.

Assume k is true:

$\displaystyle = 11^{k + 2} + 12^{2k + 1} = 133t$

Show p(k + 1):

$\displaystyle \frac{133}{11^{k + 3} + 12^{2k + 3}}$

However, this doesn't resemble any of the previous equations. Where have I erred?

 

[pka]

Assume k is true: $\displaystyle = 11^{k + 2} + 12^{2k + 1} = 133t$
Show p(k + 1):
$\displaystyle 11^{\left( {K + 1} \right) + 2} + 12^{2\left( {K + 1} \right) + 1} = 11^{K + 3} + 12^{2K + 3}$
$\displaystyle 11^{K + 3} + 12^{2K + 3} = 11^{K + 3} + 11\left( {12^{2K + 1} } \right) - 11\left( {12^{2K + 1} } \right) + 12^{2K + 3}$
\(\displaystyle 11^{\left( {K + 1} \right) + 2} + 12^{2\left( {K + 1} \right) + 1} = 11\left[ \underbrace {{11^{K + 2} + \left( {12^{2K + 1} } \right)}}_{\mbox{true for K}} \right] - \left( {12^{2K + 1} } \right)\left[ {11 - 12^2 } \right]\)

Can you finish?

 

[soroban]

Hello, Vertciel!

Your last statement is upside-down . . .

$\displaystyle \text{Prove that }E(n) \:= \:11^{n + 2} + 12^{2n + 1}\,\text{ is divisible by 133 for all positive integers }n.$

\(\displaystyle \text{Verify }E(1)\!:\;\;11^3 + 12^3 \:=\:3059 \;=\;(23)(133)\quad\hdots\;\text{True!}\)

$\displaystyle \text{Assume }E(k)\!:\;\;11^{k+2} + 12^{2k+1} \;=\;133a\,\text{ for some integer }a.$


$\displaystyle \text{Consider: }\;11^{k+3} + 12^{2k+3} \;= \;11\!\cdot\!11^{k+2} + 144\!\cdot\!12^{2k+1}$

$\displaystyle \text{Add and subtract }133\!\cdot\!11^{k+2}\!:\;\;11\!\cdot\!11^{k+2} {\bf+ 133\!\cdot\!11^{k+2}} + 144\!\cdot\!12^{2k+1} {\bf- 133\!\cdot\!11^{k+2}}$

. . . $\displaystyle = \;144\!\cdot\!11^{k+2} + 144\!\cdot\!12^{2k+1} - 133\!\cdot\!11^{k+2}$

. . . $\displaystyle =\;144\underbrace{\left(11^{k+2} + 12^{2k+1}\right)}_{\text{This is }133a} - 133\!\cdot\!11^{k+2}$

. . . $\displaystyle \;=\;144(133a) - 133\!\cdot\!11^{k+2}$

. . . \(\displaystyle =\;\underbrace{133\left(144a - 11^{k+2}\right)}_{\text{a multiple of 133}\)


$\displaystyle \text{Therefore: }\;11^{k+3} + 12^{2k+3} \;=\;133b$


\(\displaystyle \text{We have proved }E(k\!+\!1)\quad\hdots\quad\text{The inductive proof is complete.}\)