Question

3/1*4+3/4*7+3/7*10+...+3/(3x+1)*(3x+4)=2010/2011. Va rog frumos ajutati-ma, nu stiu cum sa rezolv. Va rog ajutati-ma!!

Answer 1

   
[tex] \displaystyle \\ \frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+\cdots+\frac{3}{(3x+1)(3x+4)}=\frac{2010}{2011} \\ \\ \texttt{Aplicam formula: }~~~ \boxed{\frac{3}{n(n+3)} = \frac{1}{n}-\frac{1} {n+3} } \\ \\ \frac{1}{1}-\frac{1} {4} +\frac{1}{4}-\frac{1} {7} +\frac{1}{7}-\frac{1} {10} +\cdots + \frac{1}{3x+1}-\frac{1}{3x+4}= \frac{2010}{2011} \\ \\ \frac{1}{1}-\frac{1}{3x+4}= \frac{2010}{2011} \\ \\ \frac{3x+4}{3x+4} -\frac{1}{3x+4}= \frac{2010}{2011} [/tex]


[tex]\displaystyle \\ \frac{3x+3}{3x+4}=\frac{2010}{2011} \\ \\ 2011(3x+3) = 2010(3x+4) \\ \\ 2011\cdot 3x + 2011\cdot 3 = 2010 \cdot 3x + 2010 \cdot 4 \\ \\ 2011\cdot 3x -2010 \cdot 3x = 2010 \cdot 4 -2011\cdot 3 \\ \\ 3x(2011 - 2010) = 8040 - 6033 \\ \\ 3x = 2007 \\ \\ x = \frac{2007}{3} = \boxed{\boxed{669 }}[/tex]