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 Evaluate this sum (Posted on 2005-12-21)
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              ```

 See The Solution Submitted by pcbouhid Rating: 4.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): Using Charlie's idea (spoiler?) -- numerical verification | Comment 5 of 14 |
(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//271500883549110806181523021073557232644958692985408472639853172714914204355012570197269080379120331097288409258884569210 1//271500883549110806181523021073557232644958692985408472639853172714914204355012570197269080379120331097288409258884569210 1//18870458820594643657669988986347247124415493907815750537289587528225065182046422851580018974400419291396564774701744257329251473731770450082126992655008437053281787239968163557400451486725949548946331444988517848234171391268540901362570738810 1//18870458820594643657669988986347247124415493907815750537289587528225065182046422851580018974400419291396564774701744257329251473731770450082126992655008437053281787239968163557400451486725949548946331444988517848234171391268540901362570738810 1//182320238643076193131328004973079388781938367570182620971990689317421406551399048452145355927150379502182014192105212479545276882125078544229175457511137298493257689743292275849026865210916213030064364277854824848045606897225423834150692040232702505673414251329612419495085398466108685685866598408450164161316673734102895792062757353021853638404696188190113350547238585462655629485074639831902863274162753650594202277255846136233841315380496349754491468602904919041150786650733561382010 1//182320238643076193131328004973079388781938367570182620971990689317421406551399048452145355927150379502182014192105212479545276882125078544229175457511137298493257689743292275849026865210916213030064364277854824848045606897225423834150692040232702505673414251329612419495085398466108685685866598408450164161316673734102895792062757353021853638404696188190113350547238585462655629485074639831902863274162753650594202277255846136233841315380496349754491468602904919041150786650733561382010`

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

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