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.


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




$$S=a(1+y)^n+b(1+wy)^n+(1+w^2y)^n$$ 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



Solve for $a,b$

Now find $S$ and compare with given expression


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.