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Binomial theorem

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In elementary algebra, the binomial theorem is the identity that states that for any non-negative integer n,

(x + y)^n = \sum_{k=0}^n {n \choose k} x^k y^{n-k},

or, equivalently,

(x + y)^n = \sum_{k=0}^n {n \choose k} y^k x^{n-k},

where

{n \choose k} = \frac{n!}{k!(n - k)!}.

One way to prove this identity is by mathematical induction.

Proof:

Base case: n = 0

(x + y)^0 = \sum_{k=0}^0 {0 \choose k} x^{0-k} y^k = 1

Induction case: Now suppose that it is true for n : (x + y)^n = \sum_{k=0}^n {n \choose k} x^{n-k} y^k, and prove it for n + 1.

(x+y)^{n+1} = (x+y)(x+y)^n \,
= (x+y) \sum_{k=0}^n {n \choose k} x^{n-k} y^k \,
= \sum_{k=0}^n {n \choose k} x^{n+1-k} y^k + \sum_{j=0}^n {n \choose j} x^{n-j} y^{j+1} \,
= \sum_{k=0}^n {n \choose k} x^{n+1-k} y^k + \sum_{j=0}^n {n \choose {(j+1) -1}} x^{n-j} y^{j+1} \,
= \sum_{k=0}^n {n \choose k} x^{n+1-k} y^k + \sum_{k=1}^{n+1} {n \choose {k -1}} x^{n+1-k} y^k \,
= \sum_{k=0}^{n+1} {n \choose k} x^{n+1-k} y^k - {n \choose {n+1}} x^0 y^{n+1}+ \sum_{k=0}^{n+1} {n \choose {k -1}} x^{n+1-k} y^k - {n \choose {-1}}x^{n+1} y^0 \,
= \sum_{k=0}^{n+1} \left[ {n \choose k} + {n \choose {k -1}} \right] x^{n+1-k} y^k \,
= \sum_{k=0}^{n+1} {{n+1} \choose k} x^{n+1-k} y^k,

and the proof is complete.

The first several cases

\begin{align} (x + y)^0 &= 1 \\ (x + y)^1 &= x + y \\ (x + y)^2 &= x^2 + 2xy + y^2 \\ (x + y)^3 &= x^3 + 3x^2 y + 3xy^2 + y^3 \\ (x + y)^4 &= x^4 + 4x^3 y + 6x^2 y^2 + 4xy^3 + y^4 \\ (x + y)^5 &= x^5 + 5x^4 y + 10x^3 y^2 + 6x^2 y^3 + y^5 \\ (x + y)^6 &= x^6 + 6x^5 y + 15x^4 y^2 + 20x^3 y^3 + 15 x^2 y^4 + 6xy^5 + y^6 \end{align}

Newton's binomial theorem

There is also Newton's binomial theorem, proved by Isaac Newton, that goes beyond elementary algebra into mathematical analysis, which expands the same sum (x + y)n as an infinite series when n is not an integer or is not positive.

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