# To find the sum: $\frac {1}{n!} \sum \binom {n}{2+3r} x^{1+r}$

Sum the series: $$\frac {x}{2!(n-2)!}+\frac {x^2}{5!(n-5)!}+\frac {x^3}{8!(n-8)!}+....+\frac {x^{\frac {n}{3}}}{(n-1)!},$$ $$n$$ being a multiple of $$3$$.(Math. Tripos, 1899)

# My attempt

We may write for an integer $$m$$, $$n=3m$$ so that: $$S(x)=\frac {x}{2!(n-2)!}+\frac {x^2}{5!(n-5)!}+\frac {x^3}{8!(n-8)!}+....+\frac {x^{\frac {n}{3}}}{(n-1)!}$$ $$=\frac {x}{2!(3m-2)!}+\frac {x^2}{5!(3m-5)!}+\frac {x^3}{8!(3m-8)!}+....+\frac {x^m}{(3m-1)!}$$ $$=\frac {1}{(3m)!}\left\{ \binom {3m}{2}x+\binom {3m}{5}x^2+\binom {3m}{8}x^3+....+\binom {3m}{3m-1}x^m \right\}$$ Here it is plain to me that it is some binomial expansion in which every $$2$$ out of $$3$$ terms cancel out. The first guess in such a situation is a series consisting the cube roots of unity as well as $$\sqrtx.$$ $$(\omega_3+\sqrtx)^{3m}=1+\binom {3m}{1}\omega_3^2\sqrtx+\binom {3m}{2}\omega_3\sqrt{x^2}+....+x^m$$ $$S(x)=\frac {\omega_3^2 \sqrtx}{(3m)!}\left\{ \binom {3m}{2}\omega_3\sqrt{x^2}+\binom {3m}{5}\omega_3x^2\sqrt{x^2}+....+\binom {3m}{3m-1}\omega_3x^{m-1}\sqrt{x^2}\right\}$$ Now if, $$S(x)=\frac {\omega_3^2 \sqrtx}{(3m)!} (\omega_3+\sqrtx)^{3m},$$ then we are left with two conditions: $$\binom {3m}{1}+\binom {3m}{4}x+\binom {3m}{7}x^2+....=0,$$ $$1+\binom {3m}{3}x+\binom {3m}{6}x^2+....=0$$ It seems that the whole line of logic is circular. Any help would be greatly appreciated.

## 2 Answers

This is series multisection. I'll write $$n=3m$$, like you, but consider instead $$S=\sum_{k=0}^{m-1}\binom{3m}{3k+2}y^{3k+2}.$$ Let $$S_0=\sum_{j=0}^{3m}\binom{3m}jy^j=(1+y)^{3m},$$ $$S_1=\sum_{j=0}^{3m}\binom{3m}j\omega^jy^j=(1+\omega y)^{3m}$$ and $$S_2=\sum_{j=0}^{3m}\binom{3m}j\omega^{2j}y^j=(1+\omega^2 y)^{3m}$$ where $$\omega=\exp(2\pi i/3)$$. Then $$S_0+\omega S_1+\omega^2 S_2=\sum_{j=0}^{3m}\binom{3m}j (1+\omega^{j+1}+\omega^{2j+2})y^j =3\sum_{k=0}^{m-1}\binom{3m}{3k+2}y^{3k+2}.$$ Therefore $$S=\frac{(1+y)^{3m}+\omega(1+\omega y)^{3m}+\omega^2(1+\omega^2y)^{3m}}3.$$

A change of variable gives the solution to the Tripos problem as $$\frac{\sqrtx}{n!} \frac{(1+\sqrtx)^n+\omega(1+\omega \sqrtx)^n+\omega^2(1+\omega^2\sqrtx)^n}3.$$

Hint

Consider

$$S=a(1+y)^n+b(1+wy)^n+(1+w^2y)^n$$ where $$w$$ is a complex cube (as we need every third term) root of unity

We need the coefficients of $$y^0,y^1$$ to be zero

$$a+b+1=0$$

$$an+bnw+w^2n=0$$

Solve for $$a,b$$

Now find $$S$$ and compare with given expression