Sum of nCr/(r+1) From r=0 to n

A general binomial-coefficient summation technique question, not tied to a specific exam paper.

Problem

Find:

$$\sum_{r=0}^{n} \frac{{}^{n}C_r}{r+1}$$

Key insight. Dividing a binomial coefficient by $(r+1)$ isn’t just an awkward fraction — it’s secretly a different binomial coefficient in disguise, one order up. Spotting that identity turns the whole sum into something that collapses using the standard “sum of a row of Pascal’s triangle” result.

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Approach

Rewrite $\dfrac{{}^{n}C_r}{r+1}$ as a multiple of ${}^{n+1}C_{r+1}$ using the factorial definition directly. Once every term in the sum is written this way, the whole thing becomes a sum over binomial coefficients of a fixed row — which is just $2^{n+1}$ with one term removed.

Solution

Step 1 — Rewrite the general term

Starting from the factorial definition:

$$\frac{{}^{n}C_r}{r+1} = \frac{1}{r+1} \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{(r+1)!\,(n-r)!}$$

Multiplying and dividing by $(n+1)$:

$$= \frac{1}{n+1} \cdot \frac{(n+1)!}{(r+1)!\,(n-r)!} = \frac{1}{n+1} \cdot {}^{n+1}C_{r+1}$$

since $(n-r) = (n+1)-(r+1)$.

Step 2 — Rewrite the whole sum

$$\sum_{r=0}^{n} \frac{{}^{n}C_r}{r+1} = \frac{1}{n+1}\sum_{r=0}^{n} {}^{n+1}C_{r+1}$$

Step 3 — Shift the index

As $r$ runs from $0$ to $n$, $(r+1)$ runs from $1$ to $n+1$. So:

$$\sum_{r=0}^{n} {}^{n+1}C_{r+1} = \sum_{k=1}^{n+1} {}^{n+1}C_k$$

Step 4 — Use the full-row sum of Pascal’s triangle

The sum of all binomial coefficients in row $(n+1)$, from $k=0$ to $n+1$, is $2^{n+1}$. Removing just the $k=0$ term (which equals 1):

$$\sum_{k=1}^{n+1} {}^{n+1}C_k = 2^{n+1} – 1$$

Step 5 — Combine

$$\sum_{r=0}^{n} \frac{{}^{n}C_r}{r+1} = \frac{2^{n+1}-1}{n+1}$$

Answer

Common mistakes

• Trying to sum the fractions $\dfrac{{}^{n}C_r}{r+1}$ term by term for small $n$ and guessing a pattern, instead of converting to a clean binomial identity first. The pattern-guessing approach works but is easy to get wrong without a proof.
• Forgetting to shift the index of summation after rewriting in terms of ${}^{n+1}C_{r+1}$ — the sum still needs to correctly run over $k=1$ to $n+1$, not $0$ to $n$.

Practise next

• Find $\displaystyle\sum_{r=0}^{n} \dfrac{{}^{n}C_r}{(r+1)(r+2)}$, using the same trick of converting each term into a higher-order binomial coefficient.