The sum of the first $$50$$ terms of the series: $$3 + 7 + 13 + 21 + 31 + 43 +...$$ is

$$3 + 7 + 13 + 21 + 31 + 43 +...$$
The difference of two consecutive terms is  4,6,8,10
Since the difference of two consecutive terms is in AP, the general term will be a quadratic expression in n
So the general term is t(n) = $$an^{2}+bn+c$$
t(1)= a+b+c = 3
t(2) = 4a+2b+c = 7
t(3) = 9a+3b+c = 13
we get the values of a ,b,c as 1,1,1 respectively 
General term = $$1n^{2}+n+1$$
$$\sum1n^{2}+n+1$$ = n($$\frac{n^{2}+3n+5}{3}$$) 
Substitute n = 50
=50*885

Hence D is the correct answer.