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.

  • $\begingroup$ It will be $\sum_{k=0}^\infty \frac{1}{n-k+1}\binom{n}{k}$ ;) $\endgroup$ Sep 4, 2018 at 15:26
  • $\begingroup$ It also special case of math.stackexchange.com/a/2719631/514787. $\endgroup$
    – user514787
    Sep 4, 2018 at 16:05
  • 2
    $\begingroup$ $$\sum_{k=0}^{100}\binom{100}{100-k}\frac{1}{k+1}=\int_{0}^{1}\sum_{k=0}^{100}\binom{100}{k}x^k\,dx = \int_{0}^{1}(1+x)^{100}\,dx = \frac{2^{101}-1}{101}.$$ $\endgroup$ Sep 4, 2018 at 18:42

1 Answer 1


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 $\frac{1}{101}(2^{101}-1)$.


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