Relations between binomial coefficients

Unit 22 · Binomial Expansion

Use symmetry, Pascal's rule, the ratio of consecutive coefficients, and the substitutions $x = 1$ and $x = -1$ to prove and evaluate relations and sums of binomial coefficients.

Properties of the coefficients

From the formula ${}^nC_r = \dfrac{n!}{r!,(n-r)!}$ we get three useful results.

Symmetry: ${}^nC_r = {}^nC_{n-r}$.
Pascal's rule: ${}^nC_{r-1} + {}^nC_r = {}^{n+1}C_r$ (proved in the first lesson).
Ratio of consecutive coefficients: $$\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n!}{r!,(n-r)!}\cdot\frac{(r-1)!,(n-r+1)!}{n!} = \frac{n-r+1}{r}.$$

There is also the identity $r,{}^nC_r = n,{}^{n-1}C_{r-1}$ for $1 \le r \le n$. It follows by writing $r\cdot\dfrac{n!}{r!,(n-r)!} = \dfrac{n,(n-1)!}{(r-1)!,\bigl((n-1)-(r-1)\bigr)!}$.

Sums of binomial coefficients

The binomial theorem with $a = 1$, $b = x$ is $$(1+x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \cdots + {}^nC_nx^n.$$ This is an identity, true for every $x$, so we may substitute any value.

$x = 1$: $2^n = {}^nC_0 + {}^nC_1 + {}^nC_2 + \cdots + {}^nC_n$.
$x = -1$: $0 = {}^nC_0 - {}^nC_1 + {}^nC_2 - {}^nC_3 + \cdots$.

Adding and subtracting these two results: $${}^nC_0 + {}^nC_2 + {}^nC_4 + \cdots = {}^nC_1 + {}^nC_3 + {}^nC_5 + \cdots = 2^{n-1}.$$

Example: Evaluate ${}^{10}C_0 + {}^{10}C_2 + {}^{10}C_4 + \cdots + {}^{10}C_{10}$. This is the sum of the coefficients with even $r$, which is $2^{n-1} = 2^9 = 512$.

Example: Show that ${}^nC_1 + 2,{}^nC_2 + 3,{}^nC_3 + \cdots + n,{}^nC_n = n,2^{n-1}$. Since $r,{}^nC_r = n,{}^{n-1}C_{r-1}$, the sum is $n\left({}^{n-1}C_0 + {}^{n-1}C_1 + \cdots + {}^{n-1}C_{n-1}\right) = n,2^{n-1}$. For $n = 5$: $1\cdot5 + 2\cdot10 + 3\cdot10 + 4\cdot5 + 5\cdot1 = 80 = 5\cdot2^4$.

Choosing other values of $x$ and other expressions

Any substitution that makes the right-hand side computable works. For $(1+bx)^n$ the choice $x = 1$ gives $\sum {}^nC_rb^r = (1+b)^n$.

For a polynomial $f(x)$ with coefficients $c_0, c_1, \dots$, the sum of all the coefficients is $f(1)$. The sums of the coefficients of the even powers and of the odd powers are $$\tfrac12\bigl[f(1) + f(-1)\bigr], \qquad \tfrac12\bigl[f(1) - f(-1)\bigr].$$

Example: The sum of the coefficients in the expansion of $(1+2x)^n$ is $243$. Find $n$. Put $x = 1$: $3^n = 243 = 3^5$, so $n = 5$.

Products of expansions

Since $(1+x)^m(1+x)^n = (1+x)^{m+n}$, comparing the coefficient of $x^r$ on both sides gives $$\sum_{k=0}^{r} {}^mC_k,{}^nC_{r-k} = {}^{m+n}C_r.$$ Taking $m = n$ and $r = n$, and using ${}^nC_{n-k} = {}^nC_k$, gives $\sum_{k=0}^{n}\left({}^nC_k\right)^2 = {}^{2n}C_n$.

Example: Check $\sum_{k=0}^{3}\left({}^3C_k\right)^2 = {}^6C_3$. The left side is $1^2 + 3^2 + 3^2 + 1^2 = 20$ and ${}^6C_3 = 20$.

Exam tip: ද්විපද සංගුණකවල එකතු $(1+x)^n$ හි පහසු අගයක් ආදේශ කිරීමෙන් සාධනය කෙරේ. පිළිතුරේ "$x = 1$ ආදේශ කරමු" හෝ "$x = -1$ ආදේශ කරමු" යැයි සඳහන් කරන්න; ලකුණු හිමි වන්නේ එම ආදේශ පියවරට ය. "සංගුණකවල එකතුව" සහ "පදවල එකතුව" පටලවා නොගන්න: සංගුණකවල එකතුව $f(1)$ ය, නමුත් පදවල එකතුව $x$ මත රඳා පවතී. $r,{}^nC_r = n,{}^{n-1}C_{r-1}$ යොදා ගන්නා සාධනයක දී එය මුලින් ම ප්‍රකාශ කරන්න.