Challenging Mathematical Induction Question

Vertciel

Junior Member
Joined
May 13, 2007
Messages
78
Hello there,

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

Thank you very much.

---

1. Prove that E(n)=11n+2+122n+1\displaystyle E(n) = 11^{n + 2} + 12^{2n + 1} is divisible by 133 for all positive integers n.

Show t1\displaystyle t_1 :

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

Assume k is true:

=11k+2+122k+1=133t\displaystyle = 11^{k + 2} + 12^{2k + 1} = 133t

Show p(k + 1):

13311k+3+122k+3\displaystyle \frac{133}{11^{k + 3} + 12^{2k + 3}}

However, this doesn't resemble any of the previous equations. Where have I erred?
 
Assume k is true: =11k+2+122k+1=133t\displaystyle = 11^{k + 2} + 12^{2k + 1} = 133t
Show p(k + 1):
11(K+1)+2+122(K+1)+1=11K+3+122K+3\displaystyle 11^{\left( {K + 1} \right) + 2} + 12^{2\left( {K + 1} \right) + 1} = 11^{K + 3} + 12^{2K + 3}
11K+3+122K+3=11K+3+11(122K+1)11(122K+1)+122K+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?
 
Hello, Vertciel!

Your last statement is upside-down . . .


Prove that E(n)=11n+2+122n+1 is divisible by 133 for all positive integers n.\displaystyle \text{Prove that }E(n) \:= \:11^{n + 2} + 12^{2n + 1}\,\text{ is divisible by 133 for all positive integers }n.

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

Assume E(k) ⁣:    11k+2+122k+1  =  133a for some integer a.\displaystyle \text{Assume }E(k)\!:\;\;11^{k+2} + 12^{2k+1} \;=\;133a\,\text{ for some integer }a.


Consider:   11k+3+122k+3  =  11 ⁣ ⁣11k+2+144 ⁣ ⁣122k+1\displaystyle \text{Consider: }\;11^{k+3} + 12^{2k+3} \;= \;11\!\cdot\!11^{k+2} + 144\!\cdot\!12^{2k+1}

Add and subtract 133 ⁣ ⁣11k+2 ⁣:    11 ⁣ ⁣11k+2+133 ⁣ ⁣11k+2+144 ⁣ ⁣122k+1133 ⁣ ⁣11k+2\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}}

. . . =  144 ⁣ ⁣11k+2+144 ⁣ ⁣122k+1133 ⁣ ⁣11k+2\displaystyle = \;144\!\cdot\!11^{k+2} + 144\!\cdot\!12^{2k+1} - 133\!\cdot\!11^{k+2}

. . . =  144(11k+2+122k+1)This is 133a133 ⁣ ⁣11k+2\displaystyle =\;144\underbrace{\left(11^{k+2} + 12^{2k+1}\right)}_{\text{This is }133a} - 133\!\cdot\!11^{k+2}

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

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


Therefore:   11k+3+122k+3  =  133b\displaystyle \text{Therefore: }\;11^{k+3} + 12^{2k+3} \;=\;133b


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

 
Top