Evaluate this sum in terms of n (the number of terms):
1 2 4 8 16
S = --- + ----- + ----- + ------- + ---------- + .....
2 5 41 3,281 21,523,361
(In reply to
re: Using Charlie's idea (spoiler?) by Charlie)
Using UBASIC, we can varify this formula for the first ten terms:
10 Num=1:Den=2:N=0
15 for G=1 to 10
20 Tot=Tot+Num//Den
30 print 1-Tot
35 print 2^(N+2)//(3^(2^(N+1))-1)
40 Num=Num*2:N=N+1
50 X=Num//(1-Tot)
60 Den=X+1
70 next
produces
1//2
1//2
1//10
1//10
1//410
1//410
1//1345210
1//1345210
1//28953440450810
1//28953440450810
1//26825654846035253786389446010
1//26825654846035253786389446010
1//46055408506791340513753409614892651037805514032327504332410
1//46055408506791340513753409614892651037805514032327504332410
1//2715008835491108061815230210735572326449586929854084726398531727149142043550
12570197269080379120331097288409258884569210
1//2715008835491108061815230210735572326449586929854084726398531727149142043550
12570197269080379120331097288409258884569210
1//1887045882059464365766998898634724712441549390781575053728958752822506518204
64228515800189744004192913965647747017442573292514737317704500821269926550084370
53281787239968163557400451486725949548946331444988517848234171391268540901362570
738810
1//1887045882059464365766998898634724712441549390781575053728958752822506518204
64228515800189744004192913965647747017442573292514737317704500821269926550084370
53281787239968163557400451486725949548946331444988517848234171391268540901362570
738810
1//1823202386430761931313280049730793887819383675701826209719906893174214065513
99048452145355927150379502182014192105212479545276882125078544229175457511137298
49325768974329227584902686521091621303006436427785482484804560689722542383415069
20402327025056734142513296124194950853984661086856858665984084501641613166737341
02895792062757353021853638404696188190113350547238585462655629485074639831902863
27416275365059420227725584613623384131538049634975449146860290491904115078665073
3561382010
1//1823202386430761931313280049730793887819383675701826209719906893174214065513
99048452145355927150379502182014192105212479545276882125078544229175457511137298
49325768974329227584902686521091621303006436427785482484804560689722542383415069
20402327025056734142513296124194950853984661086856858665984084501641613166737341
02895792062757353021853638404696188190113350547238585462655629485074639831902863
27416275365059420227725584613623384131538049634975449146860290491904115078665073
3561382010
showing a match between 1 minus the sum, and the formula
2^(N+2)/(3^(2^(N+1))-1)
for n = 0 to 9
|
|
Posted by Charlie
on 2005-12-21 16:18:45 |
re(2): Using Charlie's idea (spoiler?) -- numerical verification
