Question

Considere a sequência numérica

$\mathsf{a_k=-\,\dfrac{1}{2}\,cossec\,\dfrac{1}{2}\cdot cos\!\left(k-\dfrac{1}{2}\right)}$

k natural, k >= 1.

a) Obtenha a forma mais simplificada para a lei da sequência $\mathsf{b_k}$ definida por

$\mathsf{b_k=a_{k+1}-a_k}.$

para todo k.

b) Obtenha uma fórmula fechada para a soma

S(n) = sen 1 + sen 2 + sen 3 + ... + sen(n) = $\mathsf{\displaystyle\sum_{k=1}^n sen\,k}.$

Os arcos estão em radianos.

Answer 1

Seja $\mathsf{a_{k}}$ uma sequência. Se definirmos $\mathsf{\Delta(a_{k})=a_{k+1}-a_{k}}$, então $\mathsf{\Delta(x\cdot a_{k})=x\cdot\Delta(a_{k})}$ para qualquer constante $\mathsf{x}$

a)

$b_{k}=\mathsf{\Delta(a_{k})=\Delta\bigg(-\dfrac{1}{2}cossec\big(\frac{1}{2}\big)cos\big(k-\frac{1}{2}\big)\bigg)}\\\\\\b_{k}=\mathsf{\Delta(a_{k})=-\dfrac{1}{2}cossec\big(\frac{1}{2})\cdot\Delta\big(cos(k-\frac{1}{2})\big)}\\\\\\b_{k}=\mathsf{\Delta(a_{k})=-\dfrac{1}{2}cossec(\frac{1}{2})\cdot\Delta(c_{k})}$

encontrando $\mathsf{\Delta(c_{k})}$:

$\mathsf{\Delta(c_{k}):=c_{k+1}-c_{k}}\\\\\mathsf{\Delta(c_{k})=cos([k+1]-\frac{1}{2})-cos(k-\frac{1}{2})}\\\\\mathsf{\Delta(c_{k})=cos\big(k+\frac{1}{2}\big)-cos\big(k-\frac{1}{2}\big)}$

Usaremos a fórmula de prostaférese abaixo:

$\boxed{\boxed{\mathsf{cos(a)-cos(b)=-2sen\bigg(\dfrac{a+b}{2}\bigg)sen\bigg(\dfrac{a-b}{2}\bigg)}}}$


$\mathsf{\Delta(c_{k})=-2sen\bigg(\dfrac{[k+\frac{1}{2}]+[k-\frac{1}{2}]}{2}\bigg)sen\bigg(\dfrac{[k+\frac{1}{2}]-[k-\frac{1}{2}]}{2}\bigg)}\\\\\\\mathsf{\Delta(c_{k})=-2sen(k)sen\big(\frac{1}{2}\big)}$

Então:

$\mathsf{b_{k}=-\frac{1}{2}cossec(\frac{1}{2})\Delta(c_{k})}\\\\\mathsf{b_{k}=(-\frac{1}{2})\cdot cossec(\frac{1}{2})\cdot(-2)\cdot sen(\frac{1}{2})\cdot sen(k)}\\\\\boxed{\boxed{\mathsf{b_{k}=sen(k)=\Delta(a_{k})}}}$

b)

$\displaystyle\mathsf{\sum_{k=1}^{n}sen(k)=\sum_{k=1}^{n}b_{k}}\\\\\\\mathsf{\sum_{k=1}^{n}sen(k)=\sum_{k=1}^{n}\Delta(a_{k})}\\\\\\\mathsf{\sum_{k=1}^{n}sen(k)=\sum_{k=1}^{n}(a_{k+1}-a_{k})}\\\\\\\mathsf{\sum_{k=1}^{n}sen(k)=\sum_{k=1}^{n}a_{k+1}-\sum_{k=1}^{n}a_{k}}$

Note que $\mathsf{\displaystyle\sum\limits_{k=1}^{n}a_{k+1}=a_{2}+a_{3}+...+a_{n+1}=\sum\limits_{k=2}^{n+1}a_{k}}$

$\displaystyle\mathsf{\sum_{k=1}^{n}sen(k)=\sum\limits_{k=2}^{n+1}a_{k}-\sum_{k=1}^{n}a_{k}}\\\\\\\mathsf{\sum_{k=1}^{n}sen(k)=\bigg(a_{n+1}+\sum_{k=2}^{n}a_{k}\bigg)}-\bigg(a_{1}+\sum_{k=2}^{n}a_{k}\bigg)\\\\\\\mathsf{\sum_{k=1}^{n}sen(k)=a_{n+1}-a_{1}}}\\\\\\\mathsf{\sum_{k=1}^{n}sen(k)=-\dfrac{1}{2}cossec\bigg(\frac{1}{2}\bigg)cos\bigg(n+1-\frac{1}{2}\bigg)+\frac{1}{2}cossec\bigg(\frac{1}{2}\bigg)cos\bigg(1-\frac{1}{2}\bigg)}$

$\displaystyle\mathsf{\sum_{k=1}^{n}sen(k)=\dfrac{1}{2}cossec\bigg(\dfrac{1}{2}\bigg)\bigg[-cos\bigg(n+\frac{1}{2}\bigg)+cos\bigg(\frac{1}{2}\bigg)\bigg]}\\\\\\\mathsf{\sum_{k=1}^{n}sen(k)=\dfrac{1}{2}cossec\bigg(\dfrac{1}{2}\bigg)\bigg[cos\bigg(\frac{1}{2}\bigg)-cos\bigg(n+\frac{1}{2}\bigg)\bigg]}$

Aqui ainda podemos usar novamente a fórmula de prostaférese:

$\displaystyle\mathsf{\sum_{k=1}^{n}sen(k)=\dfrac{1}{2}cossec\bigg(\dfrac{1}{2}\bigg)\cdot(-2)\cdot sen\bigg(\dfrac{\frac{1}{2}+n+\frac{1}{2}}{2}\bigg)\cdot sen\bigg(\dfrac{\frac{1}{2}-n-\frac{1}{2}}{2}\bigg)}\\\\\\\mathsf{\sum_{k=1}^{n}sen(k)=-cossec\bigg(\dfrac{1}{2}\bigg)sen\bigg(\dfrac{n+1}{2}\bigg)sen\bigg(-\dfrac{n}{2}\bigg)}$

Como $\mathsf{-sen(\alpha)=sen(-\alpha)}$, temos

$\boxed{\boxed{\displaystyle\mathsf{\sum_{k=1}^{n}sen(k)=cossec\bigg(\dfrac{1}{2}\bigg)sen\bigg(\dfrac{n+1}{2}\bigg)sen\bigg(\dfrac{n}{2}\bigg)}}}$