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Given a very large number $F$ which is a repetition of a number $N$ , $M$ number of times. Is there a way to represent that number in form of geometric series.

For eg - $F=321321321\cdots 321$ where $321$ is repeated $100$ times. Then $F=321(100100100100\cdots 1001) = 321(100^{99}+100^{98}+\cdots+100+1)$

Is there a general Formula for a given $F,N,M$ ?

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The geometric series is \begin{eqnarray*} \sum_{i=0}^M x^i=\frac{x^{M+1}-1}{x-1}. \end{eqnarray*} So the formula you seek is \begin{eqnarray*} F= N\frac{x^{M+1}-1}{x-1}. \end{eqnarray*} In your example $N=321$ , $M=99$ and $x=1000$.

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    Where is $N$ in this formula and how is $x$ determined ? – 2017-02-03
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    Sorry Sammy ... typos corrected now ? $x$ is determined by asking what do I need to multiply by to get enough zero to add on the next repetition of $N$. – 2017-02-03
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    @sammy Hint: This problem solves the $(100^{90}+100^{98}+\dots+100+1)$ problem. Now multiply in the $N$ for your desired result. – 2017-02-03
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    @SimplyBeautifulArt make your first exponent $99$ not $90$. – 2017-02-03
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    Is $x$ = $10$ power(No. of digits in integer) ? – 2017-02-03
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    @sammy Yes that is right $x=10 to the power of (# digits in 321)$ ... $1000$ in this case. – 2017-02-03
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    @DonaldSplutterwit Thanks, I got it. – 2017-02-03
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    @DonaldSplutterwit Uh, you sure? I took that directly from the problem – 2017-02-03
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    @SimplyBeautifulArt I was wrong last time ... I am right this time ... ha ha – 2017-02-03