Question

Calcule a soma a baixo:

$\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10} + ... + \frac{1}{2998 \times 3001}$


#Cálculo e explicação

Answer 1

$\mathsf{\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{2998\times3001}}$

Podemos notar que as primeiras parcelas dos denominadores formam uma P.A, bem como as últimas formam outra P.A

As progressões são, respectivamente,

(1, 4, 7, ..., 2998) e (4, 7, 10, ..., 3001)

Vamos achar o termo geral dessas progressões (ambas tem razão r = 3):

$\mathsf{a_{n}=a_{1}+(n-1)r=1+(n-1)\cdot3=1+3n-3=3n-2}\\\\\mathsf{b_{n}=a_{1}+(n-1)r=4+(n-1)\cdot3=4+3n-3=3n+1}$

Agora, vamos achar o número de termos das progressões (é o mesmo para ambas):

$\mathsf{b_{n}=3001\,\,\Longleftrightarrow\,\,3n+1=3001\,\,\Longleftrightarrow\,\,3n=3000\,\,\Longleftrightarrow\,\,n=1000}$

Portanto,

$\displaystyle\mathsf{\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{2998\times3001}=\sum_{n=1}^{1000}\dfrac{1}{a_{n}\cdot b_{n}}}\\\\\\\mathsf{\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{2998\times3001}=\sum_{n=1}^{1000}\dfrac{1}{(3n-2)(3n+1)}}$

Queremos avaliar essa soma. Vamos usar argumento de frações parciais, isto é, vamos procurar constantes reais A e B tais que

$\mathsf{\dfrac{1}{(3n-2)(3n+1)}=\dfrac{A}{3n-2}+\dfrac{B}{3n+1}}$

Somando as frações:

$\mathsf{\dfrac{1}{(3n-2)(3n+1)}=\dfrac{A(3n+1)+B(3n-2)}{(3n-2)(3n+1)}}\\\\\\\mathsf{\dfrac{1}{(3n-2)(3n+1)}=\dfrac{3An+A+3Bn-2B}{(3n-2)(3n+1)}}\\\\\\\mathsf{\dfrac{1}{(3n-2)(3n+1)}=\dfrac{(3A+3B)n+(A-2B)}{(3n-2)(3n+1)}}$

Então, devemos ter $\mathsf{(3A+3B)n+(A-2B)=1=0n+1}$

Com isso, temos que resolver o sistema:

$\begin{cases}\mathsf{3A+3B=0\,\,\Longleftrightarrow\,\,A+B=0}\\\mathsf{A-2B=1}\end{cases}$

Subtraindo as equações, obtemos

$\mathsf{(A+B)-(A-2B)=0-1\,\,\Longleftrightarrow\,\,3B=-1\,\,\Longleftrightarrow\,\,\boxed{\mathsf{B=-\dfrac{1}{3}}}}$

Substituindo na primeira, $\mathsf{A+B=0\,\,\Longleftrightarrow\,\,A=-B\,\,\Longleftrightarrow\,\,\boxed{\mathsf{A=\dfrac{1}{3}}}}$

Então:

$\mathsf{\dfrac{1}{(3n-2)(3n+1)}=\dfrac{\big(\frac{1}{3}\big)}{3n-2}-\dfrac{\big(\frac{1}{3}\big)}{3n+1}}$

Note que, se definirmos a sequência abaixo

$\mathsf{c_{n}=\dfrac{\big(\frac{1}{3}\big)}{3n-2}}$

temos que

$\mathsf{c_{n+1}=\dfrac{\big(\frac{1}{3}\big)}{3(n+1)-2}=\dfrac{\big(\frac{1}{3}\big)}{3n+3-2}=\dfrac{\big(\frac{1}{3}\big)}{3n+1}}$

Portanto, dessa forma,

$\mathsf{\dfrac{1}{(3n-2)(3n+1)}=\dfrac{\big(\frac{1}{3}\big)}{3n-2}-\dfrac{\big(\frac{1}{3}\big)}{3n+1}=c_{n}-c_{n+1}}$

Agora já somos capazes de avaliar a soma:

$\displaystyle\mathsf{\sum_{n=1}^{1000}\dfrac{1}{(3n-2)(3n+1)}=\sum_{n=1}^{1000}\big[c_{n}-c_{n+1}\big]}\\\\\\\mathsf{=\big[c_{1}-c_{2}\big]+\big[c_{2}-c_{3}\big]+\big[c_{3}-c_{4}\big]+...+\big[c_{999}-c_{1000}\big]+\big[c_{1000}-c_{1001}\big]}$

Podemos ver que todas as parcelas, com exceção da primeira e da última, se cancelam:

$\displaystyle\mathsf{\sum_{n=1}^{1000}\dfrac{1}{(3n-2)(3n+1)}=c_{1}-c_{1001}}\\\\\\\mathsf{\sum_{n=1}^{1000}\dfrac{1}{(3n-2)(3n+1)}=\dfrac{\big(\frac{1}{3}\big)}{3\cdot1-2}-\dfrac{\big(\frac{1}{3}\big)}{3\cdot1001-2}}\\\\\\\mathsf{\sum_{n=1}^{1000}\dfrac{1}{(3n-2)(3n+1)}=\dfrac{1}{3}\cdot\bigg[\dfrac{1}{3-2}-\dfrac{1}{3003-2}\bigg]}\\\\\\\mathsf{\sum_{n=1}^{1000}\dfrac{1}{(3n-2)(3n+1)}=\dfrac{1}{3}\cdot\bigg[\dfrac{1}{1}-\dfrac{1}{3001}\bigg]}$

$\displaystyle\mathsf{\sum_{n=1}^{1000}\dfrac{1}{(3n-2)(3n+1)}=\dfrac{1}{3}\cdot\dfrac{3001-1}{3001}}\\\\\\\mathsf{\sum_{n=1}^{1000}\dfrac{1}{(3n-2)(3n+1)}=\dfrac{1}{3}\cdot\dfrac{3000}{3001}}\\\\\\\boxed{\boxed{\mathsf{\sum_{n=1}^{1000}\dfrac{1}{(3n-2)(3n+1)}=\dfrac{1000}{3001}}}}$