# Laws of binomial theorem

Hello folks , i wanted to discuss some really cool derivations and formulas related to binomial theorem . There is so much material on this topic that i will have to do a part 2 next week. For now , we will be focusing on proving one formula.

The Total number of terms in expansion of

$\large (a_1+a_2+a_3+\cdots+a_{k-2}+a_{k-1}+a_k)^{n}$

is equal to $\dbinom{n+k-1}{n}$ or $\dbinom{n+k-1}{k-1}$

where $n$ represents the power of the expression

and $\quad k$ represents the numbers of terms .

Now sit back tight and get some popcorn . Let's begin with the proof !

proof: From Multinomial theorem the general term of the expansion can be written as

$\large \displaystyle \sum \dfrac{n!}{n_1 ! n_2 ! \cdots n_{k-1}! n_{k}! } \times a_{1} ^{n_1}\cdot a_{2} ^{n_2}\cdots a_{k-1} ^{n_{k-1}}\cdot a_{k} ^{n_k}$ Where

$n_1+n_2+\cdots+n_{k-1}+n_{k} = n \\ n \ge n_1,n_2,n_3\cdots ,n_{k-1},n_{k} \ge 0$ $n_1,n_2,n_3\cdots ,n_{k-1},n_{k} \in \text{ Whole numbers }$

It is important to observe that number of terms in the expansion is equal to different sets of values of $\left( n_1,n_2,n_3\cdots ,n_{k-1},n_{k} \right)$ , which satisfies the above two conditions. To calculate all the permutations . Let us denote all the possible values of the variables $n_1$,$n_2$ and so on as the powers of $x$.

$\large x^0+x^1+x^{2}+\cdots+x^{n-1}+x^{n}$

$\text{Multiply the possible values of all the variables}$

$\underbrace{(x^0+x^1+x^{2}+\cdots+x^{n} )\cdot(x^0+x^1+x^{2}+\cdots+x^{n} )\cdots (x^0+x^1+x^{2}+\cdots+x^{n} )}_{k \text{ times}}$

$= (x^0+x^1+x^{2}+\cdots+x^{n-1}+x^{n})^{k}$

We want the Sum of variables to be equal to $n$ . Which means we are trying to find coefficient of $x^{n}$ . Using formula of G.P

$(x^{0}+x^1+x^{2}+\cdots+x^{n-1}+x^{n})^{k} = \dfrac{(1-x^{n+1})^{k}}{(1-x)^{k}} \\ \quad \small\color{#3D99F6}{\text{Now Expand }(1-x^{n+1})^{k} \text{ using binomial expansion}}$

$\left( 1-\underbrace{\dbinom{n}{1}\cdot x^{n+1}+\dbinom{n}{2}\cdot x^{2(n+1)}-\cdots+\dbinom{n}{n}\cdot (-1)^{k}x^{k(n+1)}}_{\text{These terms have power } > n .\text{We can ignore them} } \right) \cdot \left( 1-x \right)^{-k}$

$\text{To find coeff. of } x^{n} \text{ in } \large \dfrac{1}{(1-x)^k}$

$\dfrac{1}{(1-x)} = \sum_{r=0}^{\infty} x^{r} \quad \small\color{#3D99F6}{\text{ 0th Derivative }} \\ (-1)\dfrac{(-1)^{1}}{(1-x)^{2}} = \sum_{r=0}^{\infty} rx^{r-1} \quad \small\color{#3D99F6}{\text{ 1st Derivative }} \\ (-1)(-2)\dfrac{(-1)^{2}}{(1-x)^{3}} = \sum_{r=0}^{\infty} r(r-1)x^{r-2} \quad \small\color{#3D99F6}{\text{ 2nd Derivative }} \\ (-1)(-2)(-3)\dfrac{(-1)^{3}}{(1-x)^{4}} = \sum_{r=0}^{\infty} r(r-1)(r-2)x^{r-3} \quad \small\color{#3D99F6}{\text{ 3rd Derivative }}$

From the pattern

$(-1)(-2)(-3)\cdots(-[k-1]) \dfrac{(-1)^{k-1}}{(1-x)^{k}} = \sum_{r=0}^{\infty} r(r-1)(r-2)\cdots(r-k)x^{r-k+1} \quad \small\color{#3D99F6}{\text{ (k-1)th Derivative }}$

$(k-1)! \dfrac{(-1)^{2(k-1)}}{(1-x)^{k}} = \sum_{r=0}^{\infty} r(r-1)(r-2)\cdots(r-k)x^{r-k+1} \\ \dfrac{1}{(1-x)^{k}} = \sum_{r=0}^{\infty} \dfrac{r(r-1)(r-2)\cdots(r-k)}{(k-1)!}x^{r-k+1}$

$\dfrac{r(r-1)(r-2)\cdots(r-k)}{(k-1)!}$ is nothing but $\dbinom{r}{k-1}$ . To find coeff. of $x^n$ put $r=n+k-1$

Coefficient of $x^{n}$ is given by $\dbinom{n+k-1}{k-1}$ or $\dbinom{n+k-1}{n}$

$\large \text{Hence proved}$

I hope you enjoyed this note . Thank you.

Note by Sabhrant Sachan
4 years, 6 months ago

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The ideas that you expressed here can be summarized using the stars and bars approach.

Like you mentioned, we are looking for the number of non-negative integer solutions to the equation

$n_1 + n_2 + \ldots + n_k = n$

Staff - 4 years, 6 months ago

sweet

- 4 years, 5 months ago

COOL

- 4 years, 5 months ago