Best way to solve a summation with binomial coefficients in denominator apart from telecoping method

Question

The value of $\sum_{r=1}^{m}\frac{(m+1)(r-1)m^{r-1}}{r\binom{m}{r}} = \lambda$ then the correct statement is/are

(1) If $m=15$ and $\lambda$ is divided by m then the remainder is 14.

(2) If $m=7$ and $\lambda$ is divided by m then the remainder is 1.

(3) If $m=2$ then $\lambda$ is an even number.

(4) If $m=101$ then unit's digit of $\lambda$ is zero.

This problem is related to binomial coefficients and telescoping series. I find it difficult to realise how to approach these problems with $\binom{n}{r}$ in denominator of a finite sum.

1. I tried expressing it as $T(r) - T(r+1)$, but couldn't figure out how it could be done. Is telescoping the only way to approach this?

2. If no, then what are the alternative methods to solve this problem in the most elegant way?

3. If yes, then what is the best method to express it as $T(r) - T(r+1)$ (where $T(r)$ is a function of r)?

Answer 1

First split the $(r-1)$ term:$$\sum_{r=1}^m\dfrac{(m+1)(r-1)m^{r-1}}{r\binom mr}= \sum_{r=1}^m \left(\frac{(m+1)m^{r-1}}{\binom mr}- \frac{(m+1)m^{r-1}}{r\binom mr}\right) $$ Now we really really want to get rid of the r in the denominator of the second term, somehow.$$= \sum_{r=1}^m \left(\frac{(m+1)m^{r-1}}{\binom mr}- \frac{((m-r+1)+r)m^{r-1}}{r\binom mr}\right) $$ Notice that the $\dfrac{m^{r-1}}{\binom mr}$ term cancels out, leaving us with $$ \sum_{r=1}^m \left(\frac{(m)m^{r-1}}{\binom mr}- \frac{(m-r+1)m^{r-1}}{r\binom mr}\right)$$$$= \sum_{r=1}^m \left(\frac{m^{r}}{\binom mr}- \frac{(m-r+1)m^{r-1}}{r\binom mr} \right)$$ To simplify the second term, $$r\binom mr=\frac{m!r}{(m-r)!r!}=\frac{m!}{(r-1)!(m-r)!}$$$$=(m-r+1)\frac{m!}{(m-r+1)!(r-1)!}=(m-r+1)\binom{m}{r-1}$$ So the original expression becomes $$\sum_{r=1}^m\left(\frac{m^r}{\binom mr}-\frac{m^{r-1}}{\binom{m}{r-1}}\right)$$$$=m^m-1.$$