Mathematical Induction

**AW63** · December 22nd 2012, 01:40 PM

$5^{2n}-3^{n}$ 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

**HallsofIvy** · December 22nd 2012, 01:58 PM

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:
$5^2\cdot 5^{2k}- 5^2\cdot 3^k+ 5^2\cdot 3^k- 3\cdot 3^k$
$= 5^2(5^{2k}- 3^k)+ 3^k(25- 3)= 5^2(5^{2k}- 3^k})- 22(3^k)$

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

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

**Plato** · December 22nd 2012, 02:04 PM

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?

**AW63** · December 22nd 2012, 03:12 PM

Thank you very much for your help, much appreciated.

Thread: Mathematical Induction

LinkBack

Thread Tools

Display