# Mathematical Induction

• Dec 22nd 2012, 02:40 PM
AW63
Mathematical Induction
$5^{2n}-3^{n}$ is a multiple of 11 for all integers $n\geq 1$

My attempt at a solution so far (I haven't included shown the case for P(1) etc.

$P(k):5^{2k}-3^{k}=11p \\ P(k+1):5^{2k+2}-3^{k+1} \\ 5^{2}\cdot 5^{2k}-3\cdot 3^{k} \\5^{2}( 3^{k}+11p)-3\cdot3^{k}$

I am not sure if it's correct up to here, and what the next step is.
I have done the steps for P(1) to verify when n=1 (just not shown).
Thank you for any help.

P.S, I'm not used to using LaTeX
• Dec 22nd 2012, 02:58 PM
HallsofIvy
Re: Mathematical Induction
Quote:

Originally Posted by AW63
$5^{2n}-3^{n}$ is a multiple of 11 for all integers $n\geq 1$

My attempt at a solution so far (I haven't included shown the case for P(1) etc.

$P(k):5^{2k}-3^{k}=11p \\ P(k+1):5^{2k+2}-3^{k+1}\\5^{2}\cdot 5^{2k}-3\cdot 3^{k}$

At this point use the common step of adding and subtracting the same thing to "separate" the parts:
$5^2\cdot 5^{2k}- 5^2\cdot 3^k+ 5^2\cdot 3^k- 3\cdot 3^k$
$= 5^2(5^{2k}- 3^k)+ 3^k(25- 3)= 5^2(5^{2k}- 3^k})- 22(3^k)$

Quote:

$\\5^{2}( 3^{k}+11p)-3\cdot3^{k}$

I am not sure if it's correct up to here, and what the next step is.
I have done the steps for P(1) to verify when n=1 (just not shown).
Thank you for any help.

P.S, I'm not used to using LaTeX
• Dec 22nd 2012, 03:04 PM
Plato
Re: Mathematical Induction
Quote:

Originally Posted by AW63
$5^{2n}-3^{n}$ is a multiple of 11 for all integers $n\geq 1$

Assume we know that $P(k):5^{2k}-3^{k}=11p$ is true.

\begin{align*} 5^{2k+2}-3^{k+1}&= 5^{2k+2}-25\cdot 3^k + 25\cdot 3^{k}-3^{k+1}\\ &= 25(5^{2k}-3^{k})+3^k(25-3) \end{align*}

Can you finish?
• Dec 22nd 2012, 04:12 PM
AW63
Re: Mathematical Induction
Thank you very much for your help, much appreciated.