1. ## Lost again .

Find the nth term and the sum of the first n terms of the series .

1+11+111+1111+11111+...

I just see one more '1' is added . Not sure how to do it ..

THanks

2. Originally Posted by thereddevils
Find the nth term and the sum of the first n terms of the series .

1+11+111+1111+11111+...

I just see one more '1' is added . Not sure how to do it ..

THanks
Here's a start for you:

The terms of the series are $\displaystyle 1, \, 1 + 10, \, 1 + 10 + 10^2, \, 1 + 10 + 10^2 + 10^3, \, ....$

3. Hello, thereddevils!

This is not an easy problem . . .

Find the $\displaystyle n^{th}$ term and the sum of the first $\displaystyle n$ terms of the series:

. . . . . . $\displaystyle 1+11+111+1111+11111+\hdots$
The general term of this sequence is tricky to derive.

The 5th term has five 1's.
How can we "create" the number 11,111 from a formula?

We can get five 9's from: .$\displaystyle 10^5-1 \:=\:99,\!999$ . . . then divide by 9.

. . That is: .$\displaystyle \frac{10^5-1}{9} \:=\:11,\!111$

$\displaystyle \text{In general: }\:\frac{10^n-1}{9} \;=\;\underbrace{111\hdots 111}_{n\:1's}$

Therefore, the $\displaystyle n^{th}$ term is: .$\displaystyle \boxed{a_n \:=\:\frac{10^n-1}{9}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The sum of the first $\displaystyle n$ terms is: .$\displaystyle S \;=\; \sum^n_{k=1}\frac{10^k-1}{9}$

. . which equals: .$\displaystyle \frac{1}{9}\sum^n_{k=1}\left[10^k - 1\right] \;=\;\frac{1}{9}\left[\sum^n_{k=1}10^k - \sum^n_{k=1} 1 \right]$

The first sum is: .$\displaystyle \sum^n_{k=1} 10^k \;=\;10 + 10^2 + 10^3 + \hdots + 10^m$
. . . This is a geometric series whose sum is: .$\displaystyle \frac{10(10^n-1)}{9}$

The second sum is: .$\displaystyle \sum^n_{k=1}1 \:=\:n$

So we have: .$\displaystyle S \;=\;\frac{1}{9}\left[\frac{10(10^n-1)}{9} - n\right]$

. . Therefore: .$\displaystyle \boxed{S \;=\;\frac{1}{81}\left(10^{n+1} - 9n - 10\right]}$

4. Thanks a lot Soroban .