To solve : $$\frac{\binom{100}{1}}{100} + \frac{\binom{100}{2}}{99} + \frac{\binom{100}{3}}{98} +....+ \frac{\binom{100}{100}}{1} $$
I assumed the general form as $$\sum_{k=0}^\infty \frac{1}{n-k}\binom{n}{k} $$ to find a closed form and subtract between ($k=0$to$ \infty$) and ($k=101 $to$ \infty$) , am I on the right track? so I got stuck in the final to find close form. Please give me way to solve , Thank you in advance.
It will need much work to find those sums, into which you have written your first sum. You can use instead the $(x+1)^{100}= \binom{100}{0}x^{100}+\binom{100}{1}x^{99}+\cdots +\binom{100}{100}$, integrating both side, we get:$$\frac{(x+1)^{101}}{101}=\binom{100}{0}\frac{x^{101}}{101}+\binom{100}{1}\frac{x^{100}}{100}+\binom{100}{2}\frac{x^{99}}{99}+\cdots +\binom{100}{100}x\\ \text{or, }\binom{100}{1}\frac{x^{100}}{100}+\binom{100}{2}\frac{x^{99}}{99}+\cdots +\binom{100}{100}x = \frac{(x+1)^{101}}{101}-\binom{100}{0}\frac{x^{101}}{101}$$ if you put $x=1$, you will get the required sum, it will be $\frac{1}{101}(2^{101}-1)$.