Find partial sums of the series $12+105+1008+10011+\dots$

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 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

Answer 2

The series is:

$$\sum_{k=1}^n 10^k + 2 + 3(k-1) = \overbrace{\sum_{k=1}^n 10^k}^{S_1} + \overbrace{\sum_{k=1}^n 2}^{S_2} + \overbrace{\sum_{k=1}^n 3(k-1)}^{S_3} = \\ = \underbrace{\frac{10^{n+1}-1}{9}}_{S_1} + \underbrace{2n}_{S_2} + \underbrace{\frac{3n(n-1)}{2}}_{S_3} = \frac{10^{n+1}-1}{9} + \frac{n(3n+1)}{2}$$

Answer 3

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

Answer 4

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)