A rich body of research exists on Binomial Coefficients, with the concept finding its roots in Pascal's Triangle. My current investigation focuses on a Recursive Approach for generating these coefficients, employing only fundamental mathematical operations, thereby avoiding mathematical complexity.
The formula is:
$$x^{th}\text{ term's coefficient} =\frac{(x-1)^{th}\text{term's coefficient} \times (n-(x-2))}{(x-1)} $$ Where,
- "$n$" is the exponent of the binomial ∶ ${(a+b)}^n$
- "$x$" is the position of the specific term in the expansion whose coefficient needs to be determined
- "$(x-1)^{th}$ term's coefficient" is the coefficient of the previous term
Clarifications:
- The exponent of the binomial ${(a+b)}^n$ belongs to natural numbers. Or, $n\in \Bbb N$
- The first coefficient in the expansion can’t be found using this recursive formula, since it doesn’t have any preceding coefficient. However, it is well established that this coefficient invariably holds a value of $1$. So, we’ll implement that.
Example of Evaluation: Coefficients of ${(a+b)}^6$, evaluated using this approach:
$1^{st}$ coefficient $(x=1)$: $1$
$2^{nd}$ coefficient $(x=2)$: $ \frac{(x-1)^{th}\text{term's coefficient} \times (n-(x-2))}{(x-1)}=\frac{1\ \times (6-(2-2))}{2-1}=6$
$3^{rd}$ coefficient $(x=3)$: $ \frac{(x-1)^{th}\text{term's coefficient} \times (n-(x-2))}{(x-1)}=\frac{6\times(6-(3-2))}{3-1}=15$
$4^{th}$ coefficient $(x=4)$: $ \frac{(x-1)^{th}\text{term's coefficient} \times (n-(x-2))}{(x-1)}=\frac{15\times (6-(4-2))}{4-1}=20$
$5^{th}$ coefficient $(x=5)$: $ \frac{(x-1)^{th}\text{term's coefficient} \times (n-(x-2))}{(x-1)} =\frac{20 \times (6-(5-2))}{5-1}=15$
$6^{th}$ coefficient $(x=6)$: $ \frac{(x-1)^{th}\text{term's coefficient} \times (n-(x-2))}{(x-1)} =\frac{15\times (6-(6-2))}{6-1}=6$
$7^{th}$ coefficient $(x=7)$:$ \frac{(x-1)^{th}\text{term's coefficient} \times (n-(x-2))}{(x-1)} =\frac{6\ \times (6-(7-2))}{7-1}=1$
This recursive formula facilitates the efficient calculation of binomial coefficients, as exemplified by the evaluation of these coefficients: $1, 6, 15, 20, 15, 6, 1.$
In conclusion,
This research proposes a Recursive Formula for the Binomial Coefficients.
While initial testing suggests the formula's accuracy, further analysis is necessary to ensure its correctness across a wider range of inputs.
I welcome mathematicians for collaborations to solidify its theoretical foundation and provide formal proof of its validity or identify any potential counterexamples. Furthermore, the applicability of this approach to exponents that are irrational or fractional could also be a topic to discover.
Please let me know if you can help me with the proof.
Author:
Shaikh Sadi
$9$th Grade Student
Dhaka, Bangladesh
Let me start by introducing a more conventional way of expressing your formula.
Conventionally, we can discuss the binomial coefficients $\binom nr$ as the coefficient of $x^r$ in the expansion of $$(1+x)^n= \binom n0 x^0 + \binom n1 x^1 +... + \binom nn x^n$$
Which can be written more compactly using summation notation (and avoiding "...", which many math people object to)
$$(1+x)^n= \sum_{r=0}^n\binom nr x^r$$
A couple of things to point out here:
$$(a+b)^n = a^n\left (1+\frac ba \right )^n= \sum_{r=0}^n \binom nr a^{n-r}b^r$$
Using these conventions, your recursive formula can be written as ... $$ \binom n r = \binom n {r-1} \frac {n-(r-1)}r , (r \ge 1)$$
Now to establish the connection between binomial coefficients and combinations, consider... $$ (1+x)^3 = \color{green}{(1+x)} \color{red}{(1+x)} \color{blue}{(1+x)} \\= \color{green}{1}\cdot \color{red}{1} \cdot \color{blue}{1} +\color{green}{1}\cdot \color{red}{1} \cdot \color{blue}{x} +\color{green}{1}\cdot \color{red}{x} \cdot \color{blue}{1} +\color{green}{x}\cdot \color{red}{1} \cdot \color{blue}{1} +\color{green}{1}\cdot \color{red}{x} \cdot \color{blue}{x} +\color{green}{x}\cdot \color{red}{1} \cdot \color{blue}{x} +\color{green}{x}\cdot \color{red}{x} \cdot \color{blue}{1} +\color{green}{x}\cdot \color{red}{x} \cdot \color{blue}{x} $$
so we have $2^3=8$ terms, each obtained by choosing either a $1$ or an $x$ from each of the three brackets. The coefficient of $x^2$ in this expansion is the number of combinations of two out of the three brackets from which an $x$ is chosen, it being understood that you choose the $1$ from the remaining bracket. So $\binom 32 = 3$ because there are three combinations of two brackets that can contribute a factor of $x$, namely $\{1, 2\}, \{1, 3\}, \{2, 3\} $.
nCr(n, r), check out thatnCr(6, 3)=20Now we can develop your formula using the example $n=6, r=5$, so calculate $\binom 65$ given that $\binom 64 = 15$.
For every combination of $(r-1)=4$ objects taken from 6, we can create $n-(r-1)=2$ combinations of $r=5$ objects merely by appending one of the 2 objects not included in the combination of 4 objects. For example, one of the 4-combos is $\{2, 3, 5, 6\}$, we can create 2 5-combos using the remaining objects 1 and 4, namely... $$ \{2, 3, 5, 6\}\cup \{1\} \text{ and } \{2, 3, 5, 6\}\cup \{4\} $$
In this way you can generate every possible 5-combo, but each one will occur $r=5$ times , for instance, the single 5-combo $\{1, 2, 3, 5, 6\}$ will be generated in $r=5$ ways... $$ \{2, 3, 5, 6\}\cup \{1\} \text{ and } \{1, 3, 5, 6\}\cup \{2\} \text{ and } \{1, 2, 5, 6\}\cup \{3\} \\ \text{ and } \{1, 2, 3, 6\}\cup \{5\} \text{ and } \{1,2, 3, 5\}\cup \{6\} $$ So for each of the $\binom n{r-1}=\binom 64=15$ 4-combos we can generate $n-(r-1)= 2$ 5-combos and each 5-combo will appear $r=5$ times So... $$ \binom 65 = \binom 64 \times \frac 25 = 15 \times \frac 25 = 6 $$
and in general... $$ \binom nr = \binom n{r-1}\times \frac {n-(r-1)}r $$