The correct option is D Tn=3n−1+n
The sequence of first consecutive differences is 3,7,19,55⋯
The sequence of the second consecutive differences is 4,12,36,⋯
Clearly, it is a G.P. with common ratio 3.
The nth term of the given series is
Tn=a(3)n−1+bn2+cn+d
Putting n=1,2,3,4, we get
2=a+b+c+d5=3a+4b+2c+d12=9a+9b+3c+d31=27a+16b+4c+d
After, solving these equations, we get
a=1,b=0,c=1,d=0
Putting the values of a,b,c,d in Eq. (i), we get
Tn=3n−1+n