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: 
The series is:

$$\sum\_{k=1}^n 10^k + 2 + 3(k-1) = \overbrace{\sum\_{k=1}^n 10^k}^{S\_1} + \overbrace{\sum\_{k=1}^n 2}^{S\_2} + \overbrace{\sum\_{k=1}^n 3(k-1)}^{S\_3} = \\ = \underbrace{\frac{10^{n+1}-1}{9}}\_{S\_1} + \underbrace{2n}\_{S\_2} + \underbrace{\frac{3n(n-1)}{2}}\_{S\_3} = \frac{10^{n+1}-1}{9} + \frac{n(3n+1)}{2}$$
Well the given series can be expressed as a sum of two series one is an A.P and the other is a G.P.
 For instance,

12. = 10¹ + 2
105= 10² + 5
1008 = 10³ + 8
Therefore : first series is a G.P ie 10 + 10² + 10³+ ....+ 10^n
 Second series is an A.P 2+5 + 8 +.....+k(let)

Hence the sum of the given series is the sim of the A.P and the G.P.

Sum of G.P = a(r^n-1)/r-1 where a is the first term and 
 r is the common ratio
In this case both a and r is 10 
Therefore the sum of series is : 10(10^n -1)/9 ........(1)

Sum of A.P = (n/2)\*{2a + (n-1)d} where a is the first term 
 and d is the common difference
Therefore sum of the A.P is : (n/2)(1+3n) ...........(2)

Now if we add (1) and (2) we get the sum of the given series.
(1)+(2)= {10^(n+1) -10}/9 + n/2(1+3n)