Question

Sa se demonstreze ca pt n* au loc egalitatile :
1/1×4 + 1/4×7 + ....+1/(3n-2)(3n+1)= n/3n+1

Answer 1

$\dfrac{1}{1\cdot 4} + \dfrac{1}{4\cdot 7} + ....+\dfrac{1}{(3n-2)(3n+1)}= \dfrac{n}{3n+1} \\ \\ \text{Demonstram prin inductie matematica: }$ 

$\begin{array}{rcl} P(1) &=& \dfrac{1}{3\cdot 1+1} = \dfrac{1}{4} \\ \\ &=& \dfrac{1}{1\cdot 4} = \dfrac{1}{4} \end{array} \quad (A)\\ \\ \\ \begin{array}{rcl}P(2) &=& \dfrac{2}{3\cdot 2+1} = \dfrac{2}{7} \\ \\ &=& \dfrac{1}{1\cdot 4} + \dfrac{1}{4\cdot 7} = \dfrac{7+1}{4\cdot 7} = \dfrac{8}{4\cdot 5} = \dfrac{2}{7}\end{array}\quad (A)$

$\begin{array}{rcl} P(k) &=& \dfrac{k}{3k+1} \\ \\ &=& \dfrac{1}{1\cdot 4} + \dfrac{1}{4\cdot 7} + ....+\dfrac{1}{(3k-2)(3k+1)}\end{array}\quad (A) \\ \\ \\ P(k+1) = \dfrac{k+1}{3(k+1)+1} = \dfrac{k+1}{3k+3+1} = \boxed{\dfrac{k+1}{3k+4}}$

$P(k+1) = \\ \\ =\dfrac{1}{1\cdot 4} + \dfrac{1}{4\cdot 7} + .... +\dfrac{1}{\Big(3(k+1)-2\Big)\Big(3(k+1)+1\Big)} \\ \\ =\dfrac{1}{1\cdot 4} + \dfrac{1}{4\cdot 7} + ....+ \dfrac{1}{(3k+1)(3k+4)} \\ \\ = \dfrac{1}{1\cdot 4} + \dfrac{1}{4\cdot 7} + ....+\dfrac{1}{\Big((3k+1)-3\Big)\Big((3k+4)-3\Big)}+ \\ \\ + \dfrac{1}{(3k+1)(3k+4)} \\ \\ = \underset{P(k)}{\underbrace{\dfrac{1}{1\cdot 4} + \dfrac{1}{4\cdot 7} + ....+\dfrac{1}{(3k-2)(3k+1)}}} + \dfrac{1}{(3k+1)(3k+4)}$

$=\dfrac{k}{3k+1} + \dfrac{1}{(3k+1)(3k+4)} = \dfrac{k(3k+4)+1}{(3k+1)(3k+4)} \\ \\ = \dfrac{k(3k+3+1)+1}{(3k+1)(3k+4)} = \dfrac{k(3k+3)+k+1}{(3k+1)(3k+4)} = \\ \\= \dfrac{k\cdot 3(k+1)+k+1}{(3k+1)(3k+4)} = \dfrac{3k(k+1)+k+1}{(3k+1)(3k+4)}= \\ \\ = \dfrac{(k+1)(3k+1)}{(3k+1)(3k+4)} = \boxed{\dfrac{k+1}{3k+4}} \quad q.e.d.$