Răspuns:
Explicație pas cu pas:
a)
[tex]\frac{1}{n(n+1)} = \frac{1+n-n}{n(n+1)} = \frac{1+n}{n(n+1)} - \frac{n}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}[/tex]
b)
[tex]\frac{k}{n(n+k)} = \frac{k+n-n}{n(n+k)} = \frac{k+n}{n(n+k)} - \frac{n}{n(n+k)} = \frac{1}{n} - \frac{k}{n+k}[/tex]
c)
aplicam formula de la b)
[tex]b= \frac{5}{3*8} + \frac{7}{8*15} + \frac{9}{15*24}= \frac{5}{3*(3+5)} + \frac{7}{8*(8+7)} + \frac{9}{15*(15+9)} =[/tex]
[tex]=\frac{1}{3} - \frac{1}{3+5} + \frac{1}{8} - \frac{1}{8+7} + \frac{1}{15} - \frac{1}{15+9} = \frac{1}{3} - \frac{1}{8} + \frac{1}{8} - \frac{1}{15} + \frac{1}{15} - \frac{1}{24} =[/tex]
[tex]=\frac{1}{3} - \frac{1}{24} =\frac{8}{24} - \frac{1}{24} = \frac{7}{24}\\[/tex]
[tex]\frac{1}{4} =\frac{6}{24} < \frac{7}{24} <\frac{8}{24} =\frac{1}{3}[/tex]
deci b apartine intervalului (1/4 , 1/3)
[tex]\it a)\ \ \dfrac{^{n+1)}1}{\ \ n}-\dfrac{^{n)}1}{\ n+1}=\dfrac{\not n+1-\not n}{n(n+1)}=\dfrac{1}{n(n+1)} \Rightarrow \dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}\\ \\ \\ b)\ \dfrac{^{n+k)}1}{\ \ n}-\dfrac{^{n)}1}{\ n+k}=\dfrac{\not n+k-\not n}{n(n+1)}=\dfrac{k}{n(n+1)} \Rightarrow \dfrac{k}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+k}\\ \\ \\ c) \ \ b = \dfrac{5}{3\cdot8}+\dfrac{7}{8\cdot15}+\dfrac{9}{15\cdot24}=\dfrac{1}{3}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{24}=[/tex]
[tex]\it =\dfrac{^{8)}1}{\ 3}-\dfrac{1}{24}=\dfrac{7}{24}[/tex]
[tex]\it \dfrac{7}{24}\in(\dfrac{1}{4},\ \dfrac{1}{3}) \Rightarrow \dfrac{^{6)}1}{4}<\dfrac{7}{24}<\dfrac{^{8)}1}{3} \Rightarrow \dfrac{6}{24}<\dfrac{7}{24}<\dfrac{8}{24}\ \ (A)[/tex]
