(a + b)*c = a*c + b*c

So

(1 + x^2)*x^n = 1*x^n + x^2 * x^n

= x^n + x^{n+2}

You're thinking too hard about it.

-Dan

Edit: Say you have Sum[x^{n + 2}, n = 0, infinity] and you want to get the exponent back to "n."

Let i = n + 2. Then n = i - 2. So:

Sum[x^{n + 2}, n = 0, infinity] = Sum[x^{i - 2 + 2}, i - 2 = 0, infinity]

= Sum[x^i, i = 2, infinity]