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

\[ (x^{0}+x^1+x^{2}+\cdots+x^{n-1}+x^{n})^{k} = \dfrac{(1-x^{n+1})^{k}}{(1-x)^{k}} \\ \quad \small\color{blue}{\text{Now Expand }(1-x^{n+1})^{k} \text{ using binomial expansion}}\]

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

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

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

From the pattern

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

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

\( \dfrac{r(r-1)(r-2)\cdots(r-k)}{(k-1)!} \) 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.

## Comments

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TopNewestThe 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

\[ n_1 + n_2 + \ldots + n_k = n \] – Calvin Lin Staff · 8 months, 3 weeks ago

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– Charlotte Milanese · 7 months, 4 weeks ago

sweetLog in to reply

COOL – Charlotte Milanese · 7 months, 4 weeks ago

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