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I have a finite summation series of identical fractions e.g $ \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{5}{3}$. Now lets say I add one to the denominator for the first two values of the series. I now have $ \frac{1}{4} + \frac{1}{4} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{2}$. Now I do it again but this time to the first three values of the series. $ \frac{1}{5} + \frac{1}{5} + \frac{1}{4} + \frac{1}{3} + \frac{1}{3} = \frac{79}{60}$. To complicate things again I add one to the denominator of the last three values of the series too. $ \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{4} + \frac{1}{4} = \frac{11}{10}$

Is there an easy direct method to figure out the final answer $\frac{11}{10}$ given that we know that we did $[2,3]$ additions to the left of the series, and $[3]$ additions to the right of the series?

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up vote 1 down vote accepted

Well, lets say we are summing $n$ fractions $\frac{p}{q}$ and add $x$ to the denominators of $m$ terms.

The sum of the original series is $s = \frac{np}{q}$.

The sum of the new series is $s' = \frac{np}{q} - \frac{mp}{q} + \frac{mp}{q + x} = (m - n)\frac{p}{q} + \frac{mp}{q + x} = \frac{m - n}{n}s + \frac{mp}{q + x}$.

Repeating in this manner will allow you to generalize this to raising the denominator of some number of terms arbitrarily many times.

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Of course I am making the assumption $m \leq n$. – Alex Becker Mar 24 '11 at 4:52
Thanks for the answer, but I don't think it quite answers the question. I'm adding one to the denominators of the series for m terms, multiple times but to the same original series, not the new one. Also, I need to take into account whether the m terms are raised from the left or the right of the series. – Projectile Fish Mar 27 '11 at 10:32
Adding multiple times to the original series is the same as adding to the original series then the new series then the new new series etc. Although I do agree a problem occurs when you add to denominators which are different, but you can break the series up to avoid this problem. – Alex Becker Mar 27 '11 at 18:43

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