Whats the logic behind the series : 0,1,7,28,84,210,426,924 and so on..
As I figured it, it is basically sum of consecutive terms of (5+r)Cr where r range from 0 to whatever term I need to find(my requirement is 999998). I need to calculate this series till 10^6 which will certainly overflow by this method. Any other logic or technique?
It seems to me that $426$ is a mistake. If we have $462$ instead of $426$, then this sequence could be the following.
$$a_n= \binom{n+4}{6}$$
From $n=1$ to $15$ it gives the following sequence (A000579):
$$0, 1, 7, 28, 84, 210, 462, 924, 1716, 3003, 5005, 8008, 12376, 18564, 27132.$$
You could find other possible solutions at OEIS.
$$a_{999998} = \binom{999998+4}{6} = 1388884722215277798611116666650000.$$
You can calculate the whole sequence in an arbitrary range for example with Maple, with this code:
seq(binomial(n+4,6),n=1..1000);
This sample code gives you the sequence in the range $n=1$ to $1000$.
Setting $f(r)=r(r-1)(r-2)(r-3)(r-4)(r-5)$, note that $$(r+5)(r+4)(r+3)(r+2)(r+1)$$$$=\frac{1}{6}\{(r+6)(r+5)(r+4)(r+3)(r+2)(r+1)-(r+5)(r+4)(r+3)(r+2)(r+1)r\}$$ $$=\frac{1}{6}\{f(r+6)-f(r+5)\}.$$
So, we have $$\begin{align}\sum_{r=0}^{n}\binom{r+5}{r}&=\sum_{r=0}^{n}\binom{r+5}{5}\\&=\sum_{r=0}^{n}\frac{(r+5)(r+4)(r+3)(r+2)(r+1)}{5!}\\&=\sum_{r=0}^{n}\frac{f(r+6)-f(r+5)}{6\times 5!}\\&=\frac{1}{6!}\sum_{r=0}^{n}\{f(r+6)-f(r+5)\}\\&=\frac{1}{720}\{(f(6)-f(5))+(f(7)-f(6))+\cdots+(f(n+6)-f(n+5))\}\\&=\frac{1}{720}(f(n+6)-f(5))\\&=\frac{(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)}{720}\\&=\binom{n+6}{6}=b_n.\end{align}$$
(Note that $b_0=1,b_1=7,b_3=84,\cdots$.)
P.S. As user153012 pointed in his/her comment, your sequence can be also represented as $$\binom{n+5}{6}\ \ \ (n=0,1,2,\cdots).$$
