what type of problem is this?
I don't know whether it is a special type. the answer is easy though.
(1 - 1/2)(1 - 1/3)(1 - 1/4) . . . (1-1/4999)(1 - 1/5000)
= 1/2 * 2/3 * 3/4 * 4/5 . . . * 4998/4999 * 4999/5000 = 1/5000
All the intervening numerators and denominators cancel.
There's a trick (true more often than not!) First note that
1 - 1/n = (n-1)/n
then cancel neighboring factors in numerator & denominator, thus:
(1 - 1/2) = 1/2
(1 - 1/2)(1 - 1/3) = (1/2)*(2/3) = 1*(2/2)*(1/3) = 1/3
(1 - 1/2)(1 - 1/3)(1 - 1/4) = (1/2)*(2/3)*(3/4) = 1*(2/2)*(3/3)*(1/4) = 1/4
(1 - 1/2)(1 - 1/3)(1 - 1/4)(1 - 1/5) = (1/2)*(2/3)*(3/4)*(4/5) = 1*(2/2)*(3/3)*(4/4)*(1/5) = 1/5
. . .
So
â[j=2,n] (1 - 1/j) = 1/n
and in particular,
(1 - 1/2)(1 - 1/3)(1 - 1/4) . . . (1 - 1/5000) = 1/5000
1/5000
1/2*2/3*...
maths problem
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Verified answer
I don't know whether it is a special type. the answer is easy though.
(1 - 1/2)(1 - 1/3)(1 - 1/4) . . . (1-1/4999)(1 - 1/5000)
= 1/2 * 2/3 * 3/4 * 4/5 . . . * 4998/4999 * 4999/5000 = 1/5000
All the intervening numerators and denominators cancel.
There's a trick (true more often than not!) First note that
1 - 1/n = (n-1)/n
then cancel neighboring factors in numerator & denominator, thus:
(1 - 1/2) = 1/2
(1 - 1/2)(1 - 1/3) = (1/2)*(2/3) = 1*(2/2)*(1/3) = 1/3
(1 - 1/2)(1 - 1/3)(1 - 1/4) = (1/2)*(2/3)*(3/4) = 1*(2/2)*(3/3)*(1/4) = 1/4
(1 - 1/2)(1 - 1/3)(1 - 1/4)(1 - 1/5) = (1/2)*(2/3)*(3/4)*(4/5) = 1*(2/2)*(3/3)*(4/4)*(1/5) = 1/5
. . .
So
â[j=2,n] (1 - 1/j) = 1/n
and in particular,
(1 - 1/2)(1 - 1/3)(1 - 1/4) . . . (1 - 1/5000) = 1/5000
1/5000
1/2*2/3*...
maths problem