Algebra
posted by dian .
Simplify:
(x1/2)(x2/2)+(x3/2)...(x100/2)=

(x1/2)(x2/2)+(x3/2)...(x100/2)
= x  1/2  x + 2/2 + x  3/2  ....  x + 100/2
= xx+xx+x....x 1/2 + 2/2  3/2  ... + 100/2
looking at the last term, we see there are 100 terms, so the x's drop out
= 1/2 + 2/2  3/2  ... + 100/2
= (1/2)(1  2 + 3  .... + 99  100)
= (1/2)[ (12) + (34) + ... + (99100]
= (1/2)[ 50(1)]
= 25
Not claiming that this is the shortest or best way of doing it, but it is easy to understand 
(x1/2)(x2/2)+(x3/2)...(x100/2)
xx+xx+...+xx  1/2 + 2/2  3/2 + ...  99/2 + 100/2
the xx+xx terms cancel out, and we are left with
1/2 (1+2 3+4 ... 99+100)
= 1/2 (1+1...+1) (50 times)
= 1/2 * 50
= 25