Explicație pas cu pas:
Salut! (*/ω＼*)
Exista o regula la astfel de sume: [tex]\frac{1}{k(k+1)}=\frac{(k+1)-k}{k(k+1)}=\frac{k+1}{b(k+1)}-\frac{k}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1} \Rightarrow \frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}[/tex]
Deci la suma ta, vom avea:
[tex]a=\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\cdots+\frac{1}{29 \cdot 30}\\\\a=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{29}-\frac{1}{30}\\\\a=\frac{1}{2} -\frac{1}{30} \\\\a=\frac{15}{30} -\frac{1}{30} \\\\a=\frac{15-1}{30}\\\\a=\frac{14}{30}\\\\\boxed{a=\frac{7}{15}}\approx0.466[/tex]
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