Find partial sums of the series $12+105+1008+10011+\dots$ 
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.
 A: The series is:
$$\sum_{k=1}^n 10^k + 2 + 3(k-1) = \overbrace{\sum_{k=1}^n 10^k}^{S_1} + \overbrace{\sum_{k=1}^n 2}^{S_2} + \overbrace{\sum_{k=1}^n 3(k-1)}^{S_3} = \\ = \underbrace{\frac{10^{n+1}-1}{9}}_{S_1} + \underbrace{2n}_{S_2} + \underbrace{\frac{3n(n-1)}{2}}_{S_3} = \frac{10^{n+1}-1}{9} + \frac{n(3n+1)}{2}$$
A: Hint: $$12=10^1+3\cdot1-1$$ $$105=10^2+3\cdot2-1$$ $$1008=10^3+3\cdot3-1$$ $$100011=10^4+3\cdot4-1$$
A: 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
A: 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,


*= 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)
