Can you solve it?

Without relying upon finding patterns, we can view the first part of the problem, $B+C+D=A$ using stars and bars. We have 3 slots (B, C, D) and A objects to distribute. We then get ${{A + 3 - 1}\choose {3-1}}={{A+2}\choose 2}$ ways given a fixed first digit. We must calculate the sum for all first digits then. As A goes from 1 to 9, we can see that A+2 goes from 3 to 11.

$\sum\limits_{i=3}^{11} {i \choose 2 }=\sum\limits_{i=2}^{11} {i \choose 2} -{ 2 \choose 2}$ $= {12 \choose 3} - 1$ using hockey stick identity, ${12\choose 3 }-1=219$

If you try it with series then it is simple . It is 3+6+10+15+21+28+36+45+55 but, how it's make up I'm not telling you now . Try yourself and finally I'll tell you . Okay , now from this series , the answer is 219 . Think yourself , how it's possible.