SOLUTION: Find the sum in the series (15/42)+(15/56)+(15/72)+(15/90)+...+(15/9900)

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: Find the sum in the series (15/42)+(15/56)+(15/72)+(15/90)+...+(15/9900)       Log On


   

Question 1099656: Find the sum in the series (15/42)+(15/56)+(15/72)+(15/90)+...+(15/9900)
Found 2 solutions by richwmiller, ikleyn:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!

You are limited to 5 problems within 24 hours.
Submit one problem at a time.
No similar or duplicate problems.
Do not post the same or similar problems in more than one category.
Do not re-post a problem before 24 hours have passed.
We are not here to do your homework for you but to show you how to do one problem of a type.
If you are a registered user, you can edit your submissions.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
1%2F42 = 1%2F%286%2A7%29 = 1%2F6-1%2F7.

1%2F56 = 1%2F%287%2A8%29 = 1%2F7-1%2F8.

1%2F72 = 1%2F%288%2A9%29 = 1%2F8-1%2F9.

1%2F90 = 1%2F%289%2A10%29 = 1%2F9-1%2F10.


All your terms are of the form 

1%2F%28k%2A%28k%2B1%29%29 = 1%2Fk+-+1%2F%28k%2B1%29     (if to forget about the common numerator "15" for a minute . . . )


the last term is

1%2F9900 = 1%2F%2899%2A100%29 1%2F99-1%2F100.


Now, if to sum up all these lines, all internal terms will cancel each other, and the sum will be simply  1%2F6+-+1%2F100.


Now recall about the numerator "15" and multiply by 15. You will get your


Answer.  The sum is equal to  15%2F6+-+15%2F100.

         You can simplify it further, if you want.

Solved.