Let p be an odd integer.
for n=1 which is odd.
Suppose is odd.
Now which by part a is odd.
So the answer on the sheet was a little off.
hi the link to this question is here http://i223.photobucket.com/albums/d...f/HPIM0535.jpg
Question 8. I think I was OK in the first part with proving 2n(2n+1) is always odd.
However I am lost with part b p^n. I'm not sure if the answer written on the scan is correct.
n=1. Then p^n = p is odd.
Suppose p^n is odd. Show p^(n+1) is odd.
p^(n+1) = (p^n)*p. But we have p^n is odd, and p is odd, and the product of two odd numbers is odd, so p^(n+1) is odd.
Let be odd.
Then is also odd.
Show that P(k) being true will cause P(k+1) to be true
If P(k) is true, then is odd,
therefore since 2n is even and odd+even=odd, so P(k+1) is true if P(k) is.
Then test for an initial value, such as n=1.
There's no need to use induction for part (a) but it can be used nevertheless.
Try to show that P(k) being true causes P(k+1) to be true
if P(k) is true
Test for an initial value