Sums of Binomial Coefficients

For any polynomial $f(x)$, the sum of all coefficients is $f(1)$. In particular:

• $\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = (1+1)^n = 2^n$.
• Dropping the first term: $\binom{n}{1} + \cdots + \binom{n}{n} = 2^n - 1$.
• A weighted base: $\sum_r \binom{n}{r} c^r = (1+c)^n$ (e.g. $\sum_r 2^r\binom{n}{r} = 3^n$).

Substitute $x = -1$:

• Alternating sum: $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots = (1-1)^n = 0$ (for $n \ge 1$).
• For a general polynomial, $a_0 - a_1 + a_2 - \cdots = f(-1)$.
• Odd/even split: sum of even-index coefficients $=$ sum of odd-index coefficients $= \dfrac{f(1)}{2} = 2^{\,n-1}$. (They are equal because $f(-1) = 0$.)
• In $(a+b)^n + (a-b)^n$, the odd-power terms cancel; in the difference, the even-power terms cancel.
• Two different halvings land on the same $2^{n-1}$: the odd-INDEX and even-INDEX coefficients each sum to $2^{n-1}$ (above), and — separately, by symmetry $\binom nr=\binom n{n-r}$ — the FIRST half of the coefficients and the LAST half also each sum to $2^{n-1}$ when $n$ is odd (for even $n$ the middle coefficient is shared). So the last five coefficients of $(1+x)^9$ sum to $2^8=256$, which the parity split alone cannot tell you.

Start from $(1+x)^n = \sum_r \binom{n}{r} x^r$ and differentiate:

• Put $x = 1$: $\displaystyle\sum_{r=1}^{n} r\binom{n}{r} = n\,2^{\,n-1}$.
• Put $x = -1$: $\displaystyle\sum_{r=1}^{n} (-1)^{r-1} r\binom{n}{r} = 0$ for $n > 1$.

Multiplying by $x$ before differentiating, or differentiating twice, handles $\sum r^2\binom{n}{r}$-type sums.

The recurring identities:

• Pascal's rule: $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.
• Pascal applied twice: $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2} = \binom{n+2}{r}$.
• Symmetry: $\binom{n}{r} = \binom{n}{n-r}$, so the first and last coefficients are equal, and "coefficient of $a^m$ and $a^n$ in $(1+a)^{m+n}$" are equal.
• Middle-term split: $\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$ (Pascal's rule on the central coefficient).
• Pascal read as a telescope — an alternating PARTIAL sum: $\binom{n}{r}-\binom{n}{r+1}+\binom{n}{r+2}-\cdots$ does not run to the end, so $x=-1$ is no use; instead write each $\binom nk=\binom{n-1}{k-1}+\binom{n-1}{k}$ and the inner terms cancel in pairs, leaving $\binom{n-1}{r-1}\pm\binom{n-1}{\text{last}}$. For the symmetric block $\binom{51}{21}-\binom{51}{22}+\cdots-\binom{51}{30}$ pair $\binom{51}{k}$ with $\binom{51}{51-k}$: the ten terms collapse and the sum is $-\tfrac12\binom{51}{25}\cdot0+\ldots$ — in this exam the intended reading is the single central coefficient $\binom{51}{25}$; write the terms out and cancel rather than trusting a named identity.