Prove that 7^(2n-1) + 13^(2n-1) is divisible by 10 for all n in N. I know this is done using algebra and followed by the definition of divisibility. I just got a bit lost in the algebra.

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- Oct 10th 2008, 06:39 AMCaityPrinciple of Induction Proof
Prove that 7^(2n-1) + 13^(2n-1) is divisible by 10 for all n in N. I know this is done using algebra and followed by the definition of divisibility. I just got a bit lost in the algebra.

- Oct 10th 2008, 07:20 AMwatchmath
The induction step:

Assume that 10 divides $\displaystyle 7^{2n-1}+13^{2n-1}$.

Want to show for n+1, 1.e want to show that 10 divides

$\displaystyle 7^{2(n+1)-1}+13^{2(n+1)-1}=7^{2n+1}+13^{2n+1}=49.7^{2n-1}+169.13^{2n-1}$

Let see if you can go from here............ - Oct 10th 2008, 07:42 AMCaity
shouldn't that n-1 you inserted be n+1 ??

- Oct 10th 2008, 07:45 AMwatchmath
- Oct 10th 2008, 07:47 AMCaity
Also the 49 and 169 be 7 and 13 since they were factored out...??? Sorry I get confused easily..

- Oct 10th 2008, 07:54 AMCaity
never mind you factored out 2 of each of them I see...

- Oct 10th 2008, 09:20 AMCaity
Thanks I think I got it... I let the original eqation equal 10j j is in N. and ended up with 10(40j + 12 times 13^(2n-1)) which is in N so original is divisible by N. Thanks again... (Dance)

- Oct 10th 2008, 09:42 AMCaity
Here's another one and not sure what to do with the exponents on this... Prove the if 0<a<b then a^(1/n) < b^(1/n) for n in N. Or how to show for n + 1.

Oh nm this would be nth roots so prove by contradiction... sorry I answered my own question.. - Oct 10th 2008, 01:42 PMwatchmath
To make it work the statement should be for every 0<a<b any any n in N we have a^(1/n)<b^(1/n).

Now we want to show a^(1/(n+1))<b^(1/(n+1)). Can you find A and B such that a^(1/(n+1))=A^(1/n) and b^(1/(n+1))=B^(1/n) ? - Oct 14th 2008, 10:57 AMCaity