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P4 help wanted - series

Can anyone do question 4a on the P4 style paper in the heinemann textbook?

It is:

u = (r + 2) ^-1 - (r + 4)^-1

prove that (sum from 1 to n of u) = (7/12) - (2n + 7) / [(n + 3)(n+4)]

Sorry for the crap formatting. Any help is much appreciated

Reply 1

evariste_3086

planetzod

sum from r=1 to n (r+2)^-1=1/3+1/4+1/5+1/6...1/(n+1)+1(n+2)
sum from r=1 to n -(r+4)^-1=-1/5-1/6.......-1/(n+1)-1/(n+2)-1/(n+3)-1/(n+4)
so sum r=1 to n (r + 2) ^-1 - (r + 4)^-1=
1/3+1/4-1/(n+3)-1/(n+4)
=7/12-{n+4+n+3/(n+1)(n+4)}
=7/12-{2n+7}/(n+1)(n+4)

Reply 2

Jonny W

(sum from 1 to n) 1/(r + 2) - 1/(r + 4)
= (sum from 1 to n) 1/(r + 2) - (sum from 1 to n) 1/(r + 4)
= (sum from 1 to n) 1/(r + 2) - (sum from 3 to n + 2) 1/(r + 2)
= (sum from 1 to 2) 1/(r + 2) - (sum from n + 1 to n + 2) 1/(r + 2)
= 1/3 + 1/4 - [1/(n + 3) + 1/(n + 4)]
= 7/12 - (2n + 7) / [(n + 3)(n + 4)]

Reply 3

Womble548

If you dont understnad this:

It is a collapsing series, the difference between two summations. To make them easier write the series out vertically.

1/3 - 1/5
+ 1/4 - 1/6
+ 1/5 - 1/7

Do this up to n (skipping when you see the pattern) and you see that most of the terms cancel out.