Kaitlynteddy26
Kaitlynteddy26 Kaitlynteddy26
  • 08-08-2017
  • Mathematics
contestada

Find the sum of the first 100 terms in the series
[tex] \frac{1}{(1*2)} + \frac{1}{(2*3)} + \frac{1}{(3*4)} + . . . \frac{1}{n*(n+1)} [/tex]

Respuesta :

caylus
caylus caylus
  • 08-08-2017
Hello,

[tex] \dfrac{1}{n} - \dfrac{1}{n+1} = \dfrac{1}{n(n+1)} \\ \dfrac{1}{1*2} = \dfrac{1}{1} - \dfrac{1}{2} \\ \dfrac{1}{2*3} = \dfrac{1}{2} - \dfrac{1}{3} \\ \dfrac{1}{3*4} = \dfrac{1}{3} - \frac{1}{4} \\ ...\\ \dfrac{1}{n*(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1} \\ [/tex]

Adding member by member, we have

[tex] \dfrac{1}{1*2} + \dfrac{1}{2*3} +\dfrac{1}{3*4} +...\dfrac{1}{n*(n+1)}=\\ \dfrac{1}{1} - \dfrac{1}{n+1} \\ = \dfrac{n}{n+1} \\ [/tex]

if n=100 sum [tex]\boxed{= \dfrac{100}{101} }[/tex]


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