# Prove (x + y)^n = sum^n_{k = 0} begin{pmatrix} n \ k end{pmatrix} x^k, y^{n-k}

## Question:

Prove {eq}(x + y)^n = \sum^n_{k = 0} \begin{pmatrix} n \\ k\end{pmatrix} x^k, y^{n-k} {/eq}

## Binomial Expansion:

The binomial expansion is used for the expansion of sum or difference of two constants when there some power is given. The expansion of a term is as follow:

{eq}{\left( {a + b} \right)^n} = \sum\limits_{k = 0}^n {{}^n{C_k}{a^k}{b^{n - k}}} {/eq}

We will prove the binomial theorem by using the induction method.

For n=1,

{eq}\begin{align*} {\left( {x + y} \right)^1} &= {}^1{C_0}{x^1} + {}^1{C_1}{y^1}\\ &= \left( {x + y} \right) \end{align*}{/eq}

Let us assume the above statement is true for n=k,

{eq}{\left( {x + y} \right)^k} = {}^k{C_0}{x^k} + {}^k{C_1}{x^{k - 1}}{y^1} + ... + {}^k{C_k}{y^k} {/eq}

Now prove for n=k+1,

{eq}\begin{align*} {\left( {x + y} \right)^{k + 1}} &= \left( {x + y} \right){\left( {x + y} \right)^k}\\ &= \left( {x + y} \right)\left( {{}^k{C_0}{x^k} + {}^k{C_1}{x^{k - 1}}{y^1} + ... + {}^k{C_k}{y^k}} \right)\\ &= x\left( {{}^k{C_0}{x^k} + {}^k{C_1}{x^{k - 1}}{y^1} + ... + {}^k{C_k}{y^k}} \right) + y\left( {{}^k{C_0}{x^k} + {}^k{C_1}{x^{k - 1}}{y^1} + ... + {}^k{C_k}{y^k}} \right)\\ &= {}^k{C_0}{x^{k + 1}} + {x^k}y\left( {{}^k{C_1} + {}^k{C_0}} \right) + {x^{k - 1}}{y^2}\left( {{}^k{C_2} + {}^k{C_1}} \right) + ... + {}^k{C_k}{b^{k + 1}} \end{align*}{/eq}

Since,

{eq}\begin{align*} {}^k{C_0} + {}^k{C_1} &= \dfrac{{k!}}{{0!\left( {k - 0} \right)!}} + \dfrac{{k!}}{{1!\left( {k - 1} \right)!}}\\ &= 1 + \dfrac{{k!}}{{\left( {k - 1} \right)!}}\\ &= \dfrac{{\left( {k - 1} \right)! + k!}}{{\left( {k - 1} \right)!}}\\ &= \dfrac{{\left( {k - 1} \right)!\left( k \right)\left( {1 + k} \right)}}{{k\left( {k - 1} \right)!}}\\ &= \dfrac{{\left( {k + 1} \right)!}}{{k!}}\\ &= {}^{k + 1}{C_1} \end{align*}{/eq}

{eq}\begin{align*} {\left( {x + y} \right)^{k + 1}} &= {}^k{C_0}{x^{k + 1}} + {x^k}y\left( {{}^k{C_1} + {}^k{C_0}} \right) + {x^{k - 1}}{y^2}\left( {{}^k{C_2} + {}^k{C_1}} \right) + ... + {}^k{C_k}{b^{k + 1}}\\ &= {}^{k + 1}{C_0}{x^{k + 1}} + {}^{k + 1}{C_1}{x^k}y + ... + {}^{k + 1}{C_{k + 1}}{y^{k + 1}} \end{align*} {/eq}

Hence the above equation is true for all positive integer.

Hence,

{eq}{\left( {x + y} \right)^n} = \sum\limits_{k = 1}^n {{}^n{C_k}{x^k}{y^{n - k}}} {/eq}