Binomial theorem

The idea comes from counting. When multiplying \(n\) factors \((x+y)\), each term arises by choosing \(x\) or \(y\) in every factor. Terms with \(y^k\) come from choosing \(y\) in exactly \(k\) of the \(n\) factors, which can be done in \(\binom{n}{k}\) ways; hence the coefficient of \(x^{n-k}y^k\) is \(\binom{n}{k}\). An alternative proof uses induction on \(n\) together with Pascal's rule.

For example, \((x+2)^4=x^4+8x^3+24x^2+32x+16\): the term with \(x^2\) is \(\binom{4}{2}x^2\cdot 2^2=6\cdot 4\,x^2\). The expansion has \(n+1\) terms and its coefficients form row \(n\) of Pascal's triangle. For \((x-y)^n\) it suffices to replace \(y\) by \(-y\), so the signs alternate.

Plugging in particular values yields useful identities. With \(x=y=1\) one gets \(\sum_k\binom{n}{k}=2^n\), and with \(x=1,\ y=-1\) one gets \(\sum_k(-1)^k\binom{n}{k}=0\) for \(n\ge 1\). Typical mistake: forgetting that \(x\) and \(y\) may carry coefficients, as in \((2x-1)^4\), where the 2 is also raised to the power.