# Math Help - summation problem

1. 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

2. 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!

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

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

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

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

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

5. ok thanks i get it now

6. now i have a new problem i dont get

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

a1 - a(n-1)

but this is wrong

7. 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

$a_n-a_0$

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

Page 2 of 2 First 12