Question

1+1/(4+1/(4+1/(4+$\alpha$)))

Answer 1

Tn=3n+5 = 8+3n-3 = 8+(n-1)*3. therefore a=8 and d=3.
Hence Sn=n/2(2a+(n-1)d) = n/2(16+3n-3) = n/2(13+3n) = (13n+3n^2)/2.

Answer 2

Seriously telling I don't know the answer.