# Summation of 'n' terms of the given expression

1. Apr 5, 2014

### smart_worker

1. The problem statement, all variables and given/known data

find the general formula to calculate the sum
2. Relevant equations

1+11+111+1111+11111+.............upto n terms

3. The attempt at a solution

100 + (101+100) + (102+101 + 100) + (103 + 102+101 + 100) + ................

==> (100+100+100+....upto n terms) + (101+101+101+....upto n-1 terms) + (102+102+102+....upto n-2 terms) + .............................................+ (10n-2 + 10n-2) + (10n-1)

==>n + (101+101+101+....upto n-1 terms) + (102+102+102+....upto n-2 terms) + ...........................+ (10n-2 + 10n-2) + (10n-1)

after this i don't know how.They seem to resemble geometric series.

2. Apr 5, 2014

### LCKurtz

Notice that the kth term of that is $\frac{10^k-1}9$. Does that help?

3. Apr 9, 2014

### smart_worker

in general how do you find the general term?

GENERAL TERM:n2 ,where n>10;nεN.

am i right?

4. Apr 9, 2014

### Staff: Mentor

Yes, that's the general term. The series can be written as a summation like so:
$$\sum_{k = 11}^{\infty}k^2$$

5. Apr 9, 2014

### Ray Vickson

This is horribly divergent. However, $\sum_{k=11}^N k^2$ is meaningful for any finite $N$.

6. Apr 9, 2014

### Staff: Mentor

In your first post, you noted that each term seems to resemble a geometric series. Not only does it seem to resemble a gerometric series. That's exactly what each of the terms is. The sum of a geometric series is (arn-a)/(r-1). In your case, a = 1 and r = 10. That's how to get the general term that LCKurtz presented.

Chet