How are you doing it?
Perhaps you are doing it the fastest way.
What is your choice of answers?
1/1*2 + 1/2*3 + 1/3*4 + . . . + 1/2000*2001 + 1/2001*2002
A. 1/2002
B. 1999/2002
C. 2001/2002
D. 1
I can do this, I want to know how to do it a faster way... thanks!!
4 answers
I just add all of them, but it is taking a long time!
ok, that is not a good idea.
How about this....
let Sum(n) denote the sum of n terms,
sum(1) = 1/2
sum(2) = 1/2 + 1/6 = 4/6 = 2/3
sum(3) = 2/3 + 1/12 = 9/12 = 3/4
sum(4) = 3/4 + 1/20 = 16/20 = 4/5
notice any pattern yet ???
let's do one more
sum(5) = 4/5 + 1/30 = 25/30 = 5/6
I am very confident that
sum(n) = n/(n+1)
and since we see that there are 2001 terms
sum(2001) = 2001/2002
How about this....
let Sum(n) denote the sum of n terms,
sum(1) = 1/2
sum(2) = 1/2 + 1/6 = 4/6 = 2/3
sum(3) = 2/3 + 1/12 = 9/12 = 3/4
sum(4) = 3/4 + 1/20 = 16/20 = 4/5
notice any pattern yet ???
let's do one more
sum(5) = 4/5 + 1/30 = 25/30 = 5/6
I am very confident that
sum(n) = n/(n+1)
and since we see that there are 2001 terms
sum(2001) = 2001/2002
Oh I get it, thanks!