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I was given the question:what is 9+99+999+9999+...+999..99(30 digits) After noticing a trend, I came with the conclusion that the answer would be 28 1's 080. Can anyone confirm my answer and give a reason as to why?

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    @Ronnie: It's not; *each* summand is equal to a power of $10$ minus 1, but the entire sum is *not* a power of 10 minus 1. In words, you would say what the argument is: Notice that $10^k$ is a $1$ followed by $k$ zeros, so $10^k - 1$ is $k$ 9's. So you can replace *each summand* with a power of 10 minus 1; then you can reorder the sum so that you add all powers of 10 first, and subtract all the 1s later; then you can figure out what the sum of the powers of 10 is; etc.2010-12-20

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Note that $\underbrace{99\cdots 9}_{k\text{ digits}} = 10^k - 1.$ So your sum is the same as $(10-1) + (10^2-1) + (10^3-1) + \cdots + (10^{30}-1),$ which is equal to $(10 + 10^2 + 10^3 + \cdots + 10^{30}) - 30.$ The first sum is easy to do, the difference is easy to do, and it gives your answer.

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    I particularly like this approach since I think it could be use for general problems with repeating digits.2010-12-19
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HINT $\ $ Exploit the linearity of $\Sigma\:$: $\rm\ \Sigma\ (f(k)+c)\ = \ \Sigma\ f(k) + \Sigma\ c\ $ to reduce to a geometric sum.