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Summation of finite series

If , obtain and simplify f(r+1)-f(r).


Hence, find .
Reply 1
Avatar for james22
james22
16
What have you tried?
Original post by cucumberpj
...


Are you studying this material as an independent candidate - you seem to be struggling with a wide range of questions so I am wondering if you have actually been taught any of this
Reply 3
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cucumberpj
OP





f(r+1)-f(r)




then how to find the summation?
(edited 9 years ago)
Reply 4
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cucumberpj
OP
Original post by TenOfThem
Are you studying this material as an independent candidate - you seem to be struggling with a wide range of questions so I am wondering if you have actually been taught any of this


I learnt these all under sequences and series.
These questions seemed tricky for me.:frown:
Original post by cucumberpj
I learnt these all under sequences and series.
These questions seemed tricky for me.:frown:


They are :smile:
Reply 6
Avatar for cucumberpj
cucumberpj
OP
Original post by TenOfThem
They are :smile:


So I hope I can seek help from you all.
I am doing these in limited sources.
:redface:
Reply 7
Avatar for james22
james22
16
Original post by cucumberpj
So I hope I can seek help from you all.
I am doing these in limited sources.
:redface:


Try doing it yourself for a few small values of n to see what is happening. There should be a lot of cancelling so if you are doing big calculations you've gone wrong.
Reply 8
Avatar for davros
davros
16
Original post by cucumberpj





f(r+1)-f(r)




then how to find the summation?


Once you've done that bit, you can immediately write down:

∑r=1n[f(r+1)−f(r)]=−2∑r=1n1r(r+1)(r+2)\displaystyle \sum_{r=1}^n [f(r+1) - f(r)] = -2\sum_{r=1}^n \dfrac{1}{r(r+1)(r+2)}

The bit on the right is -2 x the sum you're asked to find, and the bit on the left simplifies enormously - try writing it out for a few values of n if you can't see what's going on!

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