Summation of 'n' terms of the given expression

AI Thread Summary
The discussion focuses on finding a general formula for the sum of the series 1 + 11 + 111 + 1111 + ... up to n terms. Participants note that each term can be expressed as (10^k - 1)/9, suggesting a geometric series structure. The general term for another series, n^2 for n > 10, is confirmed to be correct, although the series diverges when taken to infinity. The sum of the geometric series is discussed, with the formula provided as (a * r^n - a) / (r - 1), where a = 1 and r = 10. The conversation emphasizes understanding the geometric nature of the series to derive the general term effectively.
smart_worker
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Homework Statement



find the general formula to calculate the sum

Homework Equations



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

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.
 
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smart_worker said:

Homework Statement



find the general formula to calculate the sum

Homework Equations



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

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

in general how do you find the general term?

how about this series:112+122+132+...
GENERAL TERM:n2 ,where n>10;nεN.

am i right?
 
smart_worker said:
in general how do you find the general term?

how about this series:112+122+132+...
GENERAL TERM:n2 ,where n>10;nεN.

am i right?
Yes, that's the general term. The series can be written as a summation like so:
$$ \sum_{k = 11}^{\infty}k^2$$
 
Mark44 said:
Yes, that's the general term. The series can be written as a summation like so:
$$ \sum_{k = 11}^{\infty}k^2$$

This is horribly divergent. However, ##\sum_{k=11}^N k^2## is meaningful for any finite ##N##.
 
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
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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