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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 $\sqrt[3]x.$ $$ (\omega_3+\sqrt[3]x)^{3m}=1+\binom {3m}{1}\omega_3^2\sqrt[3]x+\binom {3m}{2}\omega_3\sqrt[3]{x^2}+....+x^m $$ $$ S(x)=\frac {\omega_3^2 \sqrt[3]x}{(3m)!}\left\{ \binom {3m}{2}\omega_3\sqrt[3]{x^2}+\binom {3m}{5}\omega_3x^2\sqrt[3]{x^2}+....+\binom {3m}{3m-1}\omega_3x^{m-1}\sqrt[3]{x^2}\right\} $$ Now if, $$ S(x)=\frac {\omega_3^2 \sqrt[3]x}{(3m)!} (\omega_3+\sqrt[3]x)^{3m}, $$ 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.

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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 $$\frac{\sqrt[3]x}{n!} \frac{(1+\sqrt[3]x)^n+\omega(1+\omega \sqrt[3]x)^n+\omega^2(1+\omega^2\sqrt[3]x)^n}3.$$

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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

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