# summation problem

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• Nov 7th 2009, 06:40 PM
Sneaky
ok i have a new problem,

99
E (1/i) - (1/(i+1))
i=3

i simplify it to

E 1
----------
E i^2 + E i

how do i move on, because now i = 3 in the E
• Nov 7th 2009, 06:47 PM
VonNemo19
Quote:

Originally Posted by Sneaky
ok i have a new problem,

99
E (1/i) - (1/(i+1))
i=3

i simplify it to

E 1
----------
E i^2 + E i

how do i move on, because now i = 3 in the E

Sneaky. You don't have do do all of that! Recognize what's gonna cancel, and you will see that there will only be two terms left!
• Nov 7th 2009, 06:50 PM
Sneaky
ok i am lost again, can you show me how to do it and include the answer
• Nov 7th 2009, 06:58 PM
VonNemo19
Quote:

Originally Posted by Sneaky
ok i am lost again, can you show me how to do it and include the answer

Check it out

$\displaystyle \sum_{i=3}^{99}\left(\frac{1}{i}-\frac{1}{i+1}\right)=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)$

So, you can see all of the cancellation that is gonna occur! The only thing that'll be left standing is

$\displaystyle =\frac{1}{3}-\frac{1}{100}$
• Nov 7th 2009, 06:59 PM
Sneaky
ok thanks i get it now
• Nov 7th 2009, 07:02 PM
Sneaky
now i have a new problem i dont get

n
E (ai - ai-1)
i=1

a1 - a(n-1)

but this is wrong
• Nov 7th 2009, 07:05 PM
VonNemo19
Quote:

Originally Posted by Sneaky
now i have a new problem i dont get

n
E (ai - ai-1)
i=1

Dude! It's the same exact thing. Now it's just general. What's gonna be left over after

$\displaystyle a_n-a_0$
• Nov 7th 2009, 11:09 PM
Sneaky
i have this new problem i dont know how to solve its in a picture because its too long to type
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