SOLUTION: General term in a series is Ak = (1/k) – [1/ (k+1)], where k = 2, 3 ….100. Find the sum of all the terms.

Algebra ->  Sequences-and-series -> SOLUTION: General term in a series is Ak = (1/k) – [1/ (k+1)], where k = 2, 3 ….100. Find the sum of all the terms.      Log On


   

Question 983550: General term in a series is Ak = (1/k) – [1/ (k+1)], where k = 2, 3 ….100. Find the sum of all the terms.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
A%5B2%5D=1%2F2-1%2F3
A%5B3%5D=1%2F3-1%2F4
A%5B4%5D=1%2F4-1%2F5
.
.
.
A%5B98%5D=1%2F98-1%2F99
A%5B99%5D=1%2F99-1%2F100
A%5B100%5D=1%2F100-1%2F101
So then adding all of the terms will eliminate all of the middle terms and leave you with the first and last term.
S=1%2F2-1%2F101
S=101%2F202-2%2F202
S=99%2F202