Pure Maths---Sequence

ai) aii) by MI

aiii)
an^2 - 3 bn^2 = [a(n-1)+3b(n-1)]^2 - 3 [ a(n-1)+b(n-1)]^2
=a(n-1)^2 + 6 a(n-1) b(n-1) + 9 b(n-1)^2 - 3 a(n-1)^2 - 6 a(n-1) b(n-1) - 3 b(n-1)^2
= -2 ( a(n-1)^2- 3b(n-1)^2 )
= (-2)^n

bi)
if n is odd, an^2 < 3bn^2
(an/bn) < Sqrt[3]

bii) same as bi)

biii)
standard.
show an/bn increasing for n is odd
an/bn is decreasing is even,
a(2n+1)/b(2n+1) = [2 a(2n-1)/b(2n-1) +3 ]/[a(2n-1)/b(2n-1)+2]
a(2n+1)/b(2n+1) -a(2n-1)/b(2n-1) = [ 3 - [ a(2n-1)/b(2n-1)] ^2]/[ 2 + a(2n-1)/b(2n-1)]

for n is odd, an/bn < Sqrt[3] . => increasing
for n is even, an/bn > Sqrt[3] . =>decreasing
than by sandwich principle. proved.

c)
a(n+1)/b(n+1) = (an/bn+3)/(an/bn+1)
Abs[[a(n+1)/b(n+1)-Sqrt[3]]/ [ an/bn - Sqrt[3] ] ]= Abs[(an/bn+3)/(an/bn+1)/ [ an/bn- Sqrt[3]]]
= Abs[ (an/bn-Sqrt[3])/(an/bn+1)]
< Abs [ (an/bn-Sqrt[3]) ]