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Evaluate:$$\sum_{r=1}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$$

Using the property:$$r\binom{m}{r}=m\binom{m-1}{r-1}$$

It is same as $$\sum_{r=2}^{m} \frac{(r-1)m^{r-1}}{m\cdot\binom{m-1}{r-1}}$$

How I do now?

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  • $\begingroup$ Right from the beginning we have $\sum\limits_{r=\color{red}1}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}=\sum\limits_{r=\color{red}2}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$ $\endgroup$ Commented Jul 31, 2020 at 14:23
  • $\begingroup$ @callculus yes you are right $\endgroup$
    – user69608
    Commented Jul 31, 2020 at 14:46

1 Answer 1

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Let $$S=\sum_{r=1}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$$

Multiply both sides by $m+1$.

$$S(m+1)=\sum_{r=1}^{m} \frac{(m+1)(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$$

$$S(m+1)=\sum_{r=1}^{m} \frac{(mr-(m-r+1))m^{r-1}}{r\cdot\binom{m}{r}}$$

$$S(m+1)=\sum_{r=1}^{m} \frac{(rm^r-(m-r+1)m^{r-1})}{r\cdot\binom{m}{r}}$$

$$S(m+1)=\sum_{r=1}^{m} \frac{m^r}{\binom{m}{r}}-\frac{(m-r+1)m^{r-1}r!\cdot(m-r)!} {r\cdot m!} $$

$$S(m+1)=\sum_{r=1}^{m} \frac{m^r}{\binom{m}{r}}-\frac{m^{r-1}(m-r+1)!(r-1)!}{m!}$$

$$S(m+1)=\sum_{r=1}^{m} \frac{m^r}{\binom{m}{r}}-\frac{m^{r-1}}{\binom{m}{r-1}}$$

Now this becomes a telescoping series.

$$\boxed {S(m+1)=m^m-1}$$

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  • $\begingroup$ Does my comment answer your question? Or have I misunderstood your question? $\endgroup$ Commented Jul 31, 2020 at 14:36
  • $\begingroup$ @callculus yes you were right,how did you do then? $\endgroup$
    – user69608
    Commented Jul 31, 2020 at 14:49
  • $\begingroup$ Then I´ve done nothing. For me it looks o.k. what you´ve done. $\endgroup$ Commented Jul 31, 2020 at 15:12

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