Sum of Series as $1,(2),1,(2,2),1,(2,2,2),1,(2,2,2,2),1................$ 
The Sum of First $2015$ terms of the Series...
$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,.......................$

$\bf{My\; Try::}$ We Can Write the Given Series as $$1,(2),1,(2,2),1,(2,2,2),1,(2,2,2,2),1................$$
Let we assume that This series goes to $n-$ terms, Then we can write the sum   as
$$(1+2)+(1+2+2)+(1+2+2+2)+...............+\bf{n}$$ terms.
Now how can i Calculate $\bf{n^{th}}$  terms ans Sum of the Series.
Help me Thanks
 A: The n-th term $a_n$ is $1$ iff $n$ is a triangular number i.e. of the form $k(k+1)/2$ with $k \ge 1$, and for non-triangular subscripts $a_n=2.$
So IF you stop at some $n=k(k+1)/2$ the sum up to that point will be $2n-k.$
A: Suppose $a$ is the number of $1$'s and $b$ is the number of $2$'s. Then the sum is $a+2b.$ So you just have to figure out $a$ and $b$.
A: There are $2$'s everywhere except at positions $1$, $3$, $6$, $10$, etc., at which there are $1$'s instead.  These positions have the form $T_n=\frac{1}{2}n(n+1)$ for $n\ge 1$.  The largest $n$ for which $T_n\le 2015$ is about $\sqrt{2\cdot 2015}=\sqrt{4030}\approx 63.$  When we check, we find that $T_{62} \le 2015$ and $T_{63} > 2015$, so there are exactly $62$ ones in the first $2015$ terms of the series.  The sum is therefore $2\cdot 2015 - 62 = 3968$.
A: Looking at the series in a slight variation of your pattern:
$$1,(2,1),(2,2,1),(2,2,2,1),(2,2,2,2,1),(2,2,2,2,2,1),...$$
the group size is successive integers, and the sums of the groups are successive odd numbers, giving us a square number total at each triangular number:
$$S_{T_n} = n^2$$
$T_{63} = 63\cdot 64/2 = 63*32 = 2016$ , look at that. So $S_{2016} = 63^2 = 3969$, and we know the last term at a triangular number is $1$, so $$S_{2015} =3968$$
