By the end you can
State the binomial theorem in full and in sigma notation.
Explain why the coefficients are ${}^nC_k$.
Prove the Pascal identity ${}^nC_k = {}^{n-1}C_{k-1} + {}^{n-1}C_k$.
The theorem
So the entries of row $n$ of Pascal's triangle are ${}^nC_0, {}^nC_1, \ldots, {}^nC_n$. For example, ${}^5C_2 = 10$ is entry $2$ of row $5$.
Why the coefficients are nCk
Why it works
$(1 + x)^n$ is the product of $n$ brackets $(1 + x)(1 + x)\cdots(1 + x)$. Expanding means choosing either $1$ or $x$ from each bracket and multiplying. You get $x^k$ exactly when you choose $x$ from $k$ of the $n$ brackets, and there are ${}^nC_k$ ways to choose which $k$. So the coefficient of $x^k$ is ${}^nC_k$.
The Pascal identity
Pascal identity
For $1 \le k \le n - 1$: $${}^nC_k = {}^{n-1}C_{k-1} + {}^{n-1}C_k$$
This is the addition property of the triangle, written with combinations.
Example 1: an algebraic proof ●●● Prove the Pascal identity from the formula for ${}^nC_k$.
RHS $= \dfrac{(n - 1)!}{(k - 1)!\,(n - k)!} + \dfrac{(n - 1)!}{k!\,(n - 1 - k)!}$
$= \dfrac{(n - 1)!\,k + (n - 1)!\,(n - k)}{k!\,(n - k)!}$
Common denominator k!(n − k)!: multiply the first by k/k and the second by (n − k)/(n − k).
$= \dfrac{(n - 1)!\,n}{k!\,(n - k)!} = \dfrac{n!}{k!\,(n - k)!} = {}^nC_k = $ LHS
Think about it
A counting proof: to choose $k$ people from $n$, either Ali is chosen (then choose $k - 1$ of the other $n - 1$) or Ali is not (choose all $k$ from the other $n - 1$). Adding the two cases gives the identity.
Symmetry
${}^nC_k = {}^nC_{n-k}$, which is why each row of the triangle is symmetric.
Use the theorem to write the first three terms of $(1 + 3x)^{10}$.
$1 + {}^{10}C_1(3x) + {}^{10}C_2(3x)^2 + \cdots$
$= 1 + 30x + 405x^2 + \cdots$
$(1 + x)^n$ in sigma notation
$\displaystyle\sum_{k=0}^{n} {}^nC_k x^k$
Coefficient of $x^k$ in $(1 + x)^n$
${}^nC_k$: choose $x$ from $k$ of the $n$ brackets.
Pascal identity
${}^nC_k = {}^{n-1}C_{k-1} + {}^{n-1}C_k$
Evaluate ${}^5C_2$, entry $2$ of row $5$.
Find the coefficient of $x^2$ in $(1 + 3x)^{10}$.
${}^{10}C_2 \times 9 = 45 \times 9$.
Find the coefficient of $x$ in $(1 + 3x)^{10}$.
${}^{9}C_4 + {}^{9}C_5 = $
${}^{9}C_9$ ${}^{10}C_5$ ${}^{18}C_9$ ${}^{10}C_6$ Pascal identity with $n = 10$, $k = 5$.
The coefficient of $x^k$ in $(1 + x)^n$ is ${}^nC_k$ because:
$x$ is chosen from $k$ of the $n$ brackets $k$ is less than $n$ the triangle is symmetric $(1 + x)^n = 1 + x^n$ Evaluate ${}^{10}C_5$.
$\dfrac{10 \times 9 \times 8 \times 7 \times 6}{120}$.
${}^{20}C_{17} = $
${}^{20}C_{7}$ ${}^{17}C_{3}$ ${}^{20}C_{2}$ ${}^{20}C_{3}$ Evaluate ${}^{20}C_{17}$.
${}^{20}C_3 = \dfrac{20 \times 19 \times 18}{6}$.
In the theorem $(a + b)^n = \sum {}^nC_k a^{n-k}b^k$, the term with $k = 0$ is:
$0$ $b^n$ $a^n$ $n$ Try again
New numbers every time. Keep going until you get five in a row.
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