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Remembering that...Hello Guys can you tell me how do i go about this question:
Question
Prove by Induction that for every positive integer n.
$$3^{2n} -1$$ is divisible by 8
That is a very good DIRECT proof, but does not use induction.Remembering that...
$\displaystyle \sum_{k=0}^{n} x^{n}= \frac{x^{n+1}-1}{x-1} \implies x^{n+1}-1 = (x-1)\ \sum_{k=0}^{n} x^{n}\ (1)$
... and that...
$\displaystyle 3^{2 n +2}-1 = (3^{n+1}-1)\ (3^{n+1}+1)\ (2)$
... it is evident that the expression (2) contains both 3-1 and 3+1 so that is divisible by 8...
Kind regards
$\chi$ $\sigma$
Just asume \(\displaystyle 3^{2n}-1\) is divided by 8Hello Guys can you tell me how do i go about this question:
Question
Prove by Induction that for every positive integer n.
$$3^{2n} -1$$ is divisible by 8
Indeed, that is what I was hinting at with post #2, except I wrote "9" as "$3^2$" and "8" as "$3^2 - 1$" to make the inductive nature a bit more transparent.Just asume \(\displaystyle 3^{2n}-1\) is divided by 8
and for k+1
\(\displaystyle 3^{2(k+1)}-1\)
\(\displaystyle 3^{2k}•9-1\) and replace that -1 with +8-9
\(\displaystyle 3^{2k}•9-9+8\) factour out 9 from \(\displaystyle 3^{2k}•9-9\), can you finish?
Regards,
\(\displaystyle |\pi\rangle\)
Hi Deveno,Indeed, that is what I was hinting at with post #2, except I wrote "9" as "$3^2$" and "8" as "$3^2 - 1$" to make the inductive nature a bit more transparent.
(By the way, you can use the command \ast in latex to make an asterisk (star) for "multiply" instead of using a text bullet point. If you prefer a simple dot, the command \cdot (center dot) works well also).