Calcul de sommes binomiales

Publié le 21/02/17

(cet exercice est issu de l’oral Mines-Ponts Psi 2015)
Calculer les sommes {\displaystyle\sum_{k=0}^{\lfloor n/3\rfloor}\dbinom{n}{3k}} et {\displaystyle\sum_{k=0}^{\lfloor n/4\rfloor}\dbinom{n}{4k}}.

  1. Posons {S=\displaystyle\sum_{k=0}^{\lfloor n/3\rfloor}\dbinom{n}{3k}}, {T=\displaystyle\sum_{k=0}^{\lfloor n/3\rfloor}\dbinom{n}{3k+1}}, et {U=\displaystyle\sum_{k=0}^{\lfloor n/3\rfloor}\dbinom{n}{3k+2}}.

    On évalue {f(x)=(1+x)^n=\displaystyle\sum_{k=0}^{n}\dbinom{n}{k}x^{k}}, avec {x=1}, {x=j}, {x=j^2}.

    On obtient : {\begin{cases}2^{n}=S+T+U\\(1+j)^{n}=S+jT+j^{2}U\\1+j^{2})^{n}=S+j^{2}T+jU\end{cases}}

    En ajoutant, sachant que {1+j+j^{2}}, il vient : {3S=2^{n}+2\text{Re}((1+j)^{n})}.

    Or {1+j=\text{e}^{i\pi/3}} donc {S=\displaystyle\sum_{k=0}^{\lfloor n/3\rfloor}\dbinom{n}{3k}=\dfrac{1}{3}\biggl(2^{n}+2\cos\Bigl(\dfrac{n\pi}{3}\Bigr)\biggr)}.

  2. Soit {S=\displaystyle\sum_{k=0}^{\lfloor n/4\rfloor}\dbinom{n}{4k}}, {T=\displaystyle\sum_{k=0}^{\lfloor n/4\rfloor}\dbinom{n}{4k+1}}, {U=\displaystyle\sum_{k=0}^{\lfloor n/4\rfloor}\dbinom{n}{4k+2}}, {V=\displaystyle\sum_{k=0}^{\lfloor n/4\rfloor}\dbinom{n}{4k+3}}.

    On évalue {f(x)=(1+x)^n=\displaystyle\sum_{k=0}^{n}\dbinom{n}{k}x^{k}}, avec {x\in\{1,i,-1,-i\}}

    On trouve : {\begin{cases}2^{n}=S+T+U+V\\(1+i)^{n}=S+iT-U-iV\\0=S-T+U-V\\(1-i)^{n}=S-iT-U+iV\end{cases}}

    En ajoutant, il vient : {4S=2^{n}+2\text{Re}((1+i)^{n})}, avec {1+i=\sqrt2\text{e}^{i\pi/3}}.

    Ainsi {S=\displaystyle\sum_{k=0}^{\lfloor n/4\rfloor}\dbinom{n}{4k}=\dfrac{1}{4}\biggl(2^{n}+2\sqrt{2^{n}}\cos\Bigl(\dfrac{n\pi}{4}\Bigr)\biggr)}.