Identities in Pascal’s triangle

Prove identities by substituting values such as x = 1, −1 and 2 into the binomial expansion, and by differentiating it: row sums, alternating sums and weighted sums of coefficients.

Substituting values

The binomial theorem $(1 + x)^n = \sum {}^nC_k x^k$ is true for every $x$. Choosing a clever value of $x$ turns it into an identity about the coefficients.

Example 1●●

Show that ${}^nC_0 + {}^nC_2 + {}^nC_4 + \cdots = 2^{n-1}$ for $n \ge 1$.

Add the results for $x = 1$ and $x = -1$: the odd-numbered terms cancel and the even-numbered ones double.

$2\left({}^nC_0 + {}^nC_2 + \cdots\right) = 2^n + 0$, so the sum is $2^{n-1}$.

Differentiating first

Example 2●●●

Prove that $\displaystyle\sum_{k=1}^{n} k\,{}^nC_k = n\,2^{n-1}$.

Differentiate both sides of $(1 + x)^n = \sum_{k=0}^{n} {}^nC_k x^k$:

$n(1 + x)^{n-1} = \sum_{k=1}^{n} k\,{}^nC_k x^{k-1}$

$n\,2^{n-1} = \sum_{k=1}^{n} k\,{}^nC_k$

Now substitute x = 1.

Flashcards

$sum {}^nC_k$

$2^n$ (put $x = 1$)

$sum (-1)^k,{}^nC_k$

$0$ (put $x = -1$)

$sum {}^nC_k 2^k$

$3^n$ (put $x = 2$)

$sum k,{}^nC_k$

$n\,2^{n-1}$ (differentiate, then $x = 1$)

Practice

Find the sum of the entries in row $6$ of Pascal's triangle.
Evaluate ${}^5C_0 - {}^5C_1 + {}^5C_2 - {}^5C_3 + {}^5C_4 - {}^5C_5$.
Evaluate $\sum_{k=0}^{4} {}^4C_k\,2^k$.
Evaluate ${}^6C_0 + {}^6C_2 + {}^6C_4 + {}^6C_6$.
Evaluate $\sum_{k=1}^{5} k\,{}^5C_k$.
To prove $\sum {}^nC_k = 2^n$, substitute:
An identity with $k\,{}^nC_k$ in each term usually needs you to:
How many subsets does a set of $8$ objects have, including the empty set?
Evaluate $\sum_{k=0}^{n} {}^nC_k\,(-2)^k$ when $n$ is even.