M208 – I2, this was annoying me.

I just could not figure this particular step out in I2 for ages, so thought I’d better write something here for future reference.

  \mbox{Let P be the preposition that for all integers}\ n\geq2\ , 3^{2n}-1\mbox{ is divisible by 8}\\ \mbox{Let }n=1:\\ 3^{2\times 1}-1=3^2 -1 = 8\\ 3^{2n}-1=8\mbox{, so true for }n=1 \\ \mbox{let}\  n=k, k\geq 2\mbox{, and assume }P(k)\mbox{ is true, that is:}\\ 3^{2k}-1\mbox{ is divisible by 8}\\ \mbox{Now let }n=k+1\\ 3^{2(k+1)}-1&=3^{2k+2}-1\\ &=3^23^{2k}-1\\ &=9\times3^{2k}-1\\ &=8\times3^{2k}+(3^{2k}-1)\\ \mbox{As }P(k)\mbox{ is true, this is divisible by 8. }\therefore\ P\mbox{ is true}
I just could not see how they got from the penultimate step to the last one. Then I remember that:
 9\times3^{2k}-1=3^{2k}+3^{2k}+3^{2k}+3^{2k}+3^{2k}+3^{2k}+3^{2k}+3^{2k}+3^{2k}-1\\
which can be rewritten as:
 8\times(3^{2k})+3^{2k}-1

Hopefully writing this out will help it stick in what passes for a memory around here.

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