Pure Maths Sequence

a) (i) Let P(n) be the proposition of "an,bn >0 for any positive integer n"
When n=1, a1=1>0, b1=1>0
Therefore, P(1) is true.
Suppose P(k) is true for some positive integers k,
i.e. ak>0 and bk>0
Consider n=k+1,
a(k+1)=ak+2bk>0 (because ak, bk>0)
b(k+1)=ak+bk>0 (because ak, bk>0)
So, P(k+1) is also true.
By the principle of M.I., P(n) is true for any positive integer n.

an^2-2bn^2=(a(n-1)+2b(n-1))^2 - 2(a(n-1)+b(n-1))^2
=a(n-1)^2+4a(n-1)b(n-1)+4b(n-1)^2-2a(n-1)^2-4a(n-1)b(n-1)-2b(n-1)^2
=-(a(n-1)^2-2b(n-1)^2)
=(-1)^(n-1) * (a1-2b1)
=(-1)^n

a) (ii) an+bn(root 2)=a(n-1)+2b(n-1)+a(n-1)(root 2)+b(n-1)(root 2)
=(1+root 2)(a(n-1)+b(n-1)(root 2))
=(1+root 2)^(n-1) * (a1+b1(root 2))
=(1+root 2)^n

b) (i) U(n+1)=a(n+1) / b(n+1)=(an+2bn) / (an+bn)=(an/bn+2) / (an/bn+1)=(Un+2) / (Un+1)

b) (ii) (an+bn(root 2))(an-bn(root 2))=an^2-2bn^2
[(1+root 2)^n][an-bn(root 2)]=(-1)^n
an-bn(root 2) = [-1/(1+root 2)]^n
U(2n-1)=[(-1/(1+root 2))^(2n-1) + b(2n-1)(root 2)] / b(2n-1)
=(-1/(1+root 2))^(2n-1) / b(2n-1) + root 2
< root 2 (because (-1/(1+root 2))^(2n-1) < 0 and b(2n-1) > 0)

U(2n)=[(-1/(1+root 2))^(2n) + b(2n)(root 2)] / b(2n)
=(-1/(1+root 2))^(2n) / b(2n) + root 2
> root 2 (because (-1/(1+root 2))^(2n) > 0 and b(2n) > 0)