Question: > 
> Find the sum of $n$ terms of this series-
>  $$12+105+1008+10011+.....$$
> 
> 
> 

I did not understand that how should I proceed with this problem.

Answer: 
**Hint**: $$12=10^1+3\cdot1-1$$ $$105=10^2+3\cdot2-1$$ $$1008=10^3+3\cdot3-1$$ $$100011=10^4+3\cdot4-1$$
HINT:

Taking differences of consecutive terms $$105-12=93,1008-105=903,10011-1008=9003$$

If $T\_r$ is the $r(\ge1)$th term,

$$T\_n-T\_{n-1}=9\cdot10^{n-1}+3$$ for $n\ge2$

If $\displaystyle T\_n=U\_n+V\_n$ where $U\_n=u\_{n-1}+3\implies U\_n=3n+c$ for some arbitrary constant $c$

and $\displaystyle V\_n=V\_{n-1}+9\cdot10^{n-1},V\_n-10^n=V\_{n-1}-10^{n-1}\implies V\_n-10^n=d$ for some arbitrary constant $d$

So, we can write $\displaystyle T\_n=10^n+d+3n+c=10^n+3n+K$ where $K=c+d$

For $\displaystyle n=1,12=T\_1=10+3+K\iff K=-1$

$\displaystyle\implies T\_n=10^n+3n-1$

Hope the rest should be too tough to deal with