Too much of summation!
Find the value of:
s = 1 ∑ ∞ ( 4 s − 1 1 n = 0 ∑ ∞ ( n + 1 1 k = 0 ∑ n ( ( n k ) ( k + 1 ) 4 s − 1 ( − 1 ) k ) ) − 1 )
The answer is of the form:
B A − C π ( e E π − G e D π + F )
Find: A + B + C + D + E + F
Try more here!
The answer is 25.
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1 solution
Of all your 5 lines, I understood only "First we note that". Can you explain how you get all of these?
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I used the general expansion of zeta function to create and solve the problem.
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How is this considered basic? This should not be even a level 5 question. It should be a level 12 question.
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@Pi Han Goh – I've seen many problems of zeta function on brilliant. So I thought people would be knowing the general expansion of zeta function.
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@Aditya Kumar – They are not as complicated as the ones you've made.
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@Pi Han Goh – See that there r some genius people here who'll surely solve it. Eg parth, kartik, hasan,etc.
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@Aditya Kumar – You forgot to write my name hahahahaahahaha
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@Pi Han Goh – Yeah u r genius in number theory and geometry and of course god of logic.
@Pi Han Goh – Correct! @Aditya Kumar - I had also posted a problem (not that tough) on Zeta - On the Riemann-Zeta Function! ! It has got 400 points. Thus, comparing the difficulty level of both the problems, this problem of yours should have a rating of about 2500 points :P
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@Satyajit Mohanty – Haha let's see within a week how many people solve it.
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@Aditya Kumar – Undergraduate level students can do it for sure. Only super-smart school graders will be able to do it.
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@Satyajit Mohanty – Yeah there exist some school graders who will solve it with ease. I'm sure of it. Already there is 1 solver. Waiting for more. BTW your question is amazing but I'll take time to solve it.
@Pi Han Goh – I agree. A level 12 question. @Aditya Kumar - Please provide the complete solution. How did you get to your step 4 from step 3 ?
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@Satyajit Mohanty – As I already mentioned above it is a general expansion of the zeta function. These set of questions r meant to provide information of the functions used. Then the viewer can apply them. The tagline of my set itself says that these problems are meant to be used in harder ones. Especially those provided by parth, kartik and hasan
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@Aditya Kumar – I salute you too, Sir! _/ _
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@Satyajit Mohanty – I'm 2 years younger to u. Please don't call me sir. And your problems r anyday better than mine. U require a salute.
@Abhishek Bakshi see this and comment.
No one else correct? By the way it was a tough problem simple expansion of zeta is not solving it!LOL!
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even u did it in the same way??
Note that the expansion wasn't correct previously, as the exponent was wrong.
F i r s t w e n o t e t h a t , 4 s − 1 1 n = 0 ∑ ∞ ( n + 1 1 k = 0 ∑ n ( ( n k ) ( k + 1 ) 4 s − 1 ( − 1 ) k ) ) = ζ ( 4 s ) ∴ s = 1 ∑ ∞ ( 4 s − 1 1 n = 0 ∑ ∞ ( n + 1 1 k = 0 ∑ n ( ( n k ) ( k + 1 ) 4 s − 1 ( − 1 ) k ) ) − 1 ) = s = 1 ∑ ∞ ( ζ ( 4 s ) − 1 ) = 8 7 − 4 π ( e 2 π − 1 e 2 π + 1 )