If the series is telescoping, what is C(n) "C sub n" where
A(n) = C(n) - C(n+1)
the series is sum of (9/5) * ( 1 / ((n+1)*(n+2)) ) n = 1 to infinity
The series is telescopic, but i'm not real sure what it is asking for C(n). Any explanation would be appreciated. Thanks
What is C(n)?
it's asking you to write it in explicit terms (without the sigma sign). These are sometimes difficult but let's see
you can factor out the 9/5
so n=1 to infinity of 1/(n+1)(n+2)
ok so here the best idea is to use partial fractions
so I see that (n+2)-(n+1) = 1 so I figure the fraction must be
(n+2)-(n+1) / (n+1)(n+2)
so 1/(n+1) - 1/(n+2)
A(n)
so C(n) is 1/(n+1)
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