By the binomial theorem,
[tex]\displaystyle \left(1+\frac xk\right)^n = \sum_{i=0}^n \binom ni \left(\dfrac xk\right)^n[/tex]
The [tex]x^2[/tex] term occurs for i = 2, and the [tex]x^3[/tex] term for i = 3. If the coefficient of the squared term is 3 times larger than the coefficient of the cubed term, then
[tex]\dbinom n2 \dfrac1{k^2} = 3\dbinom n3\dfrac1{k^3}[/tex]
Solve for k :
[tex]\dbinom n2 \dfrac1{k^2} = 3 \dbinom n3 \dfrac1{k^3} \\\\ \dfrac{n!}{2!(n-2)!} \dfrac1{k^2} = \dfrac{3n!}{3!(n-3)!} \dfrac1{k^3} \\\\ \dfrac{n!}{2(n-2)(n-3)!} \dfrac1{k^2} = \dfrac{3n!}{6(n-3)!} \dfrac1{k^3} \\\\ \dfrac{1}{(n-2)k^2} = \dfrac{1}{k^3}\\\\ k^3 = (n-2)k^2 \\\\ k^3 - (n-2)k^2 = 0 \\\\ k^2 (k - (n - 2)) = 0[/tex]
Since k ≠ 0, it follows that
k - (n - 2) = 0
or
k = n - 2
as required.
