What is the value of the expression:
(1+1/2)(1+1/3)(1+1/4)...
(1+1/2004)(1+1/2005)
the dots just mean it has skipped the brackets in between.
calculators may not be used for this question.
Any help on how to solve this question? The answer is 2003 but I'd like to know how to work it out.
2007-11-06 05:53:23 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Write the factors out with common denominators, and I am sure you will see what happens.
This allows you to check your proposed answer as well.
2007-11-06 06:04:25 · answer #1 · answered by Anonymous · 1⤊ 0⤋
Notice that 1 + 1/2 = 3/2, 1 + 1/3 = 4/3, 1 + 1/4 = 5/4, ..., 1 + 1/2004 = 2005/2004, and 1 + 1/2005 = 2006/2005. Therefore, the problem is
(3/2)(4/3)(5/4) ... (2005/2004)(2006/2005). Observe that the first numerator cancels with the second denominator, 2nd num cancels 3rd denom, ... , next to last num cancels last denom. The only factors remaining after all the cancellation are the last num divided by the first denom, or 2006/2 = 1003.
2007-11-06 14:08:27 · answer #2 · answered by Tony 7 · 1⤊ 0⤋
It's 3/2 * 4/3 * 5/4 * 6/5 * ...* 2005/2004 * 2006/2005
= 2006/2 = 1003
2007-11-06 14:08:19 · answer #3 · answered by Dr D 7 · 2⤊ 0⤋
(1+1/2)(1+1/3)(1+1/4)... = (3/2)(4/3)(5/4)...=(n+2)!/(2 (n+1)!) = (n+2)/2.
So for n=2005-1=2004, we have 2006/2=1003 ! and not 2003 !!!!!!!!!!!
2007-11-06 14:07:55 · answer #4 · answered by ?????? 7 · 2⤊ 0⤋
remans sums
2007-11-06 14:02:51 · answer #5 · answered by BlueBelle84 3 · 0⤊ 1⤋