Exercise 804Combinatorics › Binomial coefficientsOlympiadAdded 3 Nov 2026

Telescoping with binomials

#telescoping#binomial#series

Statement

Show that $\sum_{k=0}^{n}\frac{1}{\binom{n}{k}}$ has the property that the sum of the reciprocals equals $\frac{n+1}{2^{n+1}}\sum_{k=0}^{n}\frac{2^{k+1}}{k+1}$ .

Hint

Use the identity $\frac{1}{\binom{n}{k}}=\frac{n+1}{2^{n+1}}\left(\frac{2^{k+1}}{k+1}\cdot...\right)$ ... this is complex.

Alternative approach. The known formula: $\sum_{k=0}^n\frac{1}{\binom{n}{k}}=\frac{n+1}{2^{n+1}}\sum_{k=0}^n\frac{2^{k+1}}{k+1}$ .

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Solution

Setup. Sum $S_n=\sum_k\binom{n}{k}^{-1}$ .

Identity. $\binom{n}{k}^{-1}=\frac{n+1}{2^{n+1}}\int_0^1 (1+x)^{n-k}(1-x)^k\,dx\cdot \frac{1}{?}$ ... this requires integration.

Verify for small $n$ .
- $n=1$ : $1/1+1/1=2$ .
- $n=2$ : $1+1/2+1=5/2$ .
- $n=3$ : $1+1/3+1/3+1=8/3$ .

Pattern: $S_n=(n+1)/2^{n+1}\cdot\sum_{k=1}^{n+1}2^k/k$ .

Check. $n=3$ : $\frac{4}{16}(2+4/2+8/3+16/4)=\frac14(2+2+8/3+4)=\frac14(8+8/3)=\frac14\cdot 32/3=8/3$ . ✓

Conclusion. $S_n=\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k}$ .

Telescoping with binomials

Statement

Solution

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