n=10
7*10-1=69!=k*6 when k is natural.
Not true... hmm, I'll include what the book says.
"In the case n = 1, 7n - 1 is divisible by 6 so P(1) is true.
Assume now the 7k - 1 is divisible by 6 for some k >= 1.
Then,
7k + 1 = 7(7k) - 1
= 7(7k - 1) + 7 -1
= 7)7k - 1) + 6
Since 7k - 1 is divisible by 6 it follows that 7(7k - 1) + 6 is also divisible by 6."
Make your post clearer next time and you will get a proper answer. As you can see, from your first post it looks like you are trying to prove that which is obviously wrong (take n=2, for example).
However I guess you are trying to prove that , which is correct. If this is the case, you have shown P(1) holds. Assume P(k), and now we need to prove that P(k+1) follows.
Note that:
Now, by the induction hypothesis (that is, ), we get:
for some .
But which is divisible by 6, therefore P(k+1) holds and we are done.
If you want to follow your book's way:
But,
Now, by the induction hypothesis, and therefore for some . Substitute and get:
and obviously therefore P(k+1) holds and we are done.