Arithmetic sequence in coefficients of $(x+y)^n$

Question

For which $n$ the second and third and fourth term's coefficients of binomial expansion's $(x+y)^n$ makes arithmetic sequence?

Answer 1

From $2\dbinom n2=\dbinom n1+\dbinom n3$ we get $$n(n-1)=n+\frac{n(n-1)(n-2)}6$$ Since $n\neq 0$, $$6n-6=6+n^2-3n+2$$ $$n^2-9n+14=0$$ so $n=2$ or $n=7$. For $n=2$ we don't have a 'fourth term'. Thus, $n=7$.