Finding unknowns in a binomial expansion

When a question gives some terms or coefficients of an expansion such as \((1+kx)^n\), the unknown power \(n\) and the unknown constant \(k\) are found by writing the same terms with the binomial theorem and matching them.

Why: two expansions of the same expression are equal for every \(x\), so the coefficients of each power of \(x\) must be equal. Each given coefficient therefore gives one equation in the unknowns.

Method

1. Write the terms you need with the general term \(\binom nr a^{n-r}b^r\). For \((1+kx)^n\): \(1+nkx+\dfrac{n(n-1)}{2}k^2x^2+\dfrac{n(n-1)(n-2)}{6}k^3x^3+\cdots\)
2. Set each coefficient equal to its given value. For a ratio of coefficients, write the fraction and cancel the common factors of \(\binom nr\) before solving.
3. With two unknowns, make one the subject of the simpler equation (for example \(k=\dfrac{c}{n}\) from \(nk=c\)), substitute into the other and solve.
4. Reject any \(n\) that is not a positive integer, and apply any condition such as \(k>0\).
5. When the unknown appears as a common factor on both sides, as in \(pk^3=qk^4\), factorise instead of dividing: \(k^3(p-qk)=0\) keeps the case \(k=0\) unless the question excludes it.

Example

In the expansion of \((a+2x)^5\), where \(a>0\), the coefficient of \(x\) is \(160\). Find \(a\).

Coefficient of \(x\): \(\binom51a^4(2)=10a^4=160\), so \(a^4=16\) and \(a=\pm2\). Since \(a>0\), \(a=2\).

Example

In the expansion of \(\left(1+\dfrac{x}{2}\right)^n\), where \(n\ge3\), the coefficient of \(x^3\) is equal to the coefficient of \(x^2\). Find \(n\).

\(\binom n3\left(\dfrac12\right)^3=\binom n2\left(\dfrac12\right)^2\), so \(\binom n3=2\binom n2\).

Since \(n\ge3\), \(\binom n2\ne0\), so divide by it: \(\dfrac{\binom n3}{\binom n2}=\dfrac{n-2}{3}=2\), so \(n=8\).