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Question 8. I think I was OK in the first part with proving 2n(2n+1) is always odd.

This is not the product of 2 odd numbers as 2n is even.

Let \(\displaystyle n\) be odd.

Then \(\displaystyle n\pm2k\) is also odd.

**P(k)**
\(\displaystyle n\left(n\pm2k\right)=odd\)

**P(k+1)**
\(\displaystyle n\left(n\pm2(k+1)\right)=odd\)

Show that P(k) being true will cause P(k+1) to be true

**Proof**
\(\displaystyle n\left(n\pm2k\right)=n^2\pm2kn=odd\) ?

\(\displaystyle n\left(n\pm2(k+1)\right)=n^2\pm2kn\pm2n\)

If P(k) is true, then \(\displaystyle n^2\pm2kn\) 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.

Part (b)

p=odd

**P(k)**
\(\displaystyle p^k=odd\)

**P(k+1)**
\(\displaystyle p^{k+1}=odd\)

Try to show that P(k) being true causes P(k+1) to be true

**Proof**
\(\displaystyle p^{k+1}=(p)\left(p^k\right)=(odd)(odd)\) if P(k) is true

Test for an initial value