Let
pr=(1/2)*(3/4)*(5/6)... *(99/100)
Without explicit computation prove that
pr is less than 1/10, but more than 1/15.
The number is (100!/(50!*2^50)) / (50!*2^50), which reduces to 100! / ((50!)^2 * 2^100).
For what it's worth in guiding a solution to the problem, the computed value is:
12611418068195524166851562157/158456325028528675187087900672
The prime factors of the numerator (which is reduced to lowest terms) are:
3 3 3 3 11 13 17 19 29 31 53 59 61 67 71 73 79 83 89 97
The prime factors of the denominator are all 2's:
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
The value is approximately
0.0795892373871787614 or 1/12.564512901854901014
5 open "bondprod.txt" for output as #2
10 V=1
20 for I=1 to 99 step 2
30 V=V*I//(I+1)
40 next
50 print #2, V
60 N=num(V)
70 while N>1
80 Pd=prmdiv(N):N=N//Pd
90 print #2, Pd;
100 wend
110 print #2,:print #2,
160 N=den(V)
170 while N>1
180 Pd=prmdiv(N):N=N//Pd
190 print #2, Pd;
200 wend
210 print #2,:print #2,
220 print #2, V/1,1/V
|
Posted by Charlie
on 2015-02-13 10:49:00 |