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

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- Nov 7th 2009, 06:40 PMSneaky
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 PMVonNemo19
- Nov 7th 2009, 06:50 PMSneaky
ok i am lost again, can you show me how to do it and include the answer

- Nov 7th 2009, 06:58 PMVonNemo19
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 PMSneaky
ok thanks i get it now

- Nov 7th 2009, 07:02 PMSneaky
now i have a new problem i dont get

n

E (ai - ai-1)

i=1

i think the answer is

a1 - a(n-1)

but this is wrong - Nov 7th 2009, 07:05 PMVonNemo19
- Nov 7th 2009, 11:09 PMSneaky
i have this new problem i dont know how to solve its in a picture because its too long to type