Question

b) S= 1/6*7+1/7*8+1/8*9+...+1/74*75
c) S= 1/4*5+1/5*6+1/6*7+...+1/71*72

Answer 1

 

$\displaystyle\bf\\Folosim~formula:\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\\\\b)\\\\S=\frac{1}{6\times7}+\frac{1}{7\times8}+\frac{1}{8\times9}+\cdots+\frac{1}{73\times74}+\frac{1}{74\times75}\\\\S=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\cdots+\frac{1}{73}-\frac{1}{74}+\frac{1}{74}-\frac{1}{75}\\\\\text{Reducem termenii asemenea:}\\\\S=\frac{1}{6}-\frac{1}{75}=\frac{25}{6\times25}-\frac{2}{75\times2}=\frac{25}{150}-\frac{2}{150}=\boxed{\bf\frac{23}{150}}$

$\displaystyle\bf\\c)\\\\S=\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+\cdots+\frac{1}{70\times71}+\frac{1}{71\times72}\\\\S=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\cdots+\frac{1}{70}-\frac{1}{71}+\frac{1}{71}-\frac{1}{72}\\\\\text{Reducem termenii asemenea:}\\\\S=\frac{1}{4}-\frac{1}{72}=\frac{18}{4\times18}-\frac{1}{72}=\frac{18}{72}-\frac{1}{72}=\boxed{\bf\frac{17}{72}}$