Question

Use indução matemática para provar que a proposição dada é verdadeira para todo inteiro positivo n.
(a) 1/1.4+1/4.7+1/7.10+⋯+1/((3n-2)(3n+1))=n/(3n+1)

Answer 1

Caso Base:

Note que a proposicao eh verdadeira para n=1:

$\dfrac{1}{(3(1)-2))(3(1)+1)}=\dfrac{1}{(1)(4)}=\dfrac14=\dfrac{(1)}{(3(1)+1)}=\dfrac14$

Passo de inducao:

Deixe k>1 e assuma que

$\sum_{i=1}^k\ \dfrac{1}{(3i-2)(3i+1)}=\dfrac{k}{3k+1}$

Portanto:

$\sum_{i=1}^{k+1}\ \dfrac{1}{(3i-2)(3i+1)}=\\\\\\\\\dfrac{1}{(3(k+1)-2)(3(k+1)+1)}+\sum_{i=1}^{k}\ \dfrac{1}{(3i-2)(3i+1)}=\\\\\\\dfrac{1}{(3(k+1)-2)(3(k+1)+1)}+\dfrac{k}{3k+1}=\\\\\\\dfrac{1}{(3k+1)(3k+4)}+\dfrac{k}{3k+1}=\\\\\\\dfrac{1+k(3k+4)}{(3k+1)(3k+4)}=\\\\\\\dfrac{3k^2+4k+1}{(3k+1)(3k+4)}=\\\\\\\dfrac{(3k+1)(k+1)}{(3k+1)(3k+4)}=\\\\\\\dfrac{k+1}{3k+4}=\\\\\\\dfrac{k+1}{3(k+1)+1}$

Portanto, por inducao

$\sum_{i=1}^n\ \dfrac{1}{(3i-2)(3i+1)}=\dfrac{n}{3n+1}$

Q.E.D