Question

need to find the equations of the following, using Ó (and the n=1 which is at the bottom of the Ó) (and infinity is at the top of the Ó)

3) -6/3 - 4/4 - 2/5 - 0 + . . .

4) 9/1 + 36/4 +27/9 + 144/16 + . . .

5) 20/4 + 25/5 + 30/6 + 35/7 +40/8 + ...

i still don't grasp how to get the answer

Answer 1

Huh? Can you retype the question?

Answer 2

need to find the equations of the following, n=1 which is used to find the first answer if the series and so forth
3) -6/3 - 4/4 - 2/5 - 0 + . . .

4) 9/1 + 36/4 +27/9 + 144/16 + . . .

5) 20/4 + 25/5 + 30/6 + 35/7 +40/8 + ...

i still don't grasp how to get the answer

Answer 3

You have to look for patterns in your numbers

3) -6/3 - 4/4 - 2/5 - 0 + ...

the numerators are -6, -4, -2, 0, ...
they appear to decrease by a factor of 2

so I know it must be something like
-2n + k
what will k have to be so your first result is -6 if n=1 ?
shouldn't it be k = -4 ?

so try -2n-4, and plug in n = 1,2,3,...
what do you get?

now look at the denominator.
3, 4, 5, ...
looks like it goes up by 1
so how about n + k ?
wouldn't n+3 give you 3 when n = 1 ?
try n = 2,3,...

4) 9/1 + 36/4 + 27/9 + 144/16 + ...

This one is rather subtle until you try reducing each term, wouldn't you get 9 for each one ?

so [sigma] 9 as n=1 to infinity

let me know what you get for 5)

Answer 4

< wouldn't n+3 give you 3 when n = 1 ? >

should read:
wouldn't n+2 give you 3 when n = 1 ?