Maclaurin Series for $(\sec(x))^2$

Question

Question 2a ii

So I know I have to use the given tangent MacLaurin Series for solve for 2aii, but how did

$(\sec(x))^2 = 1+ (a_1x + a_3x^3 + a_5x^5 +….)^2 $

turn into
$ (\sec(x))^2= 1 + a_1^2x^2 + 2a_1a_3x^4 +….$ (Real Answer)

I thought it would be $a_1^2x^2 + a_3^2x^6$

Answer 1

The term $2a_1 a_3 x^4$ in (Real Answer) comes from $$(a_1x + a_3x^3 + ...)^2=(a_1x + a_3x^3 + ...)(a_1x + a_3x^3 + ...)=a_1^2 x^2+a_1x a_3x^3+a_3x^3a_1x+a_3^2 x^6=a_1^2 x^2+2a_1a_3x^4 +a_3^2 x^6$$

So what you computed is also a term in the Maclaurin series for $sec^2(x)$.