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30 Oct 2007, 08:36
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can someone explain how

h(100) + 1 = 2 ^50 * 50 ! + 1 .

where h(n) is defined as the product of all even integers from 2 to n inclusive.?
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30 Oct 2007, 08:43
h(100)=2*4*6*8.....*100
it's the same as (1*2)*(2*2)*(3*2)*.....(50*2)
simplified1*2*3*4*5*.....*50*2^50

because we have to multiply 2 by 50times so there's 2^50

h(100)+1=50!*2^50+1

hope it can help u
30 Oct 2007, 08:43
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