Question

Exercițiul 3 în care calculați N. OFER COROANA. ​

Answer 1

 

$\displaystyle\bf\\\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\\\\N=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}+\frac{2}{400}\\\\N=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}+\frac{2}{400}\\\\Se~reduc~fractiile~2~cate~2~care~au~semn~contrar.\\\\N=\frac{1}{1}-\frac{1}{200}+\frac{2}{400}\\\\N=\frac{1}{1}-\frac{1}{200}+\frac{1}{200}\\\\\boxed{\bf N=1}\\\\\boxed{\bf\sqrt{1}=\pm1}$