Instead of thinking of the sequence starting with the 1st term,
think of it starting with the 0th term:

0th term =  0 = 0×3  = 0×(0+3)
1st term =  5 = 1×5  = 1×(2+3)
2nd term = 14 = 2×7  = 2×(4+3)
3rd term = 27 = 3×9  = 3×(6+3)
4th term = 44 = 4×11 = 4×(8+3) 
5th term = 65 = 5×13 = 5×(10+3) 

So each term is the term number times twice the term number plus 3.   

So the nth term, counting starting with 0 is n(2n+3)

But if you want to start counting with 1 instead of 0,
you have to replace n by 1 less than n, so it would be

an = (n-1)[2(n-1)+3) = (n-1)[2n-2+3] = (n-1)(2n-1)

Edwin