summation problem

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

Quote:

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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

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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!

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

Quote:

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Originally Posted by Sneaky

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

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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

$\displaystyle =\frac{1}{3}-\frac{1}{100}$

ok thanks i get it now

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

Quote:

---

Originally Posted by Sneaky

now i have a new problem i dont get

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

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

$\displaystyle a_n-a_0$

i have this new problem i dont know how to solve its in a picture because its too long to type