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@Satyajit Mohanty
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I find that more troublesome, unless you got a table for the number of Bernoulli numbers, then yes, it will be easier, otherwise it takes up more time.
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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2^{34}
a_{i-1}
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Hint:
k=1∑n((k+1)2−k2)(n+1)2−1k=1∑nk===2k=1∑nk+k=1∑n12k=1∑nk+n2n(n+1)
And
k=1∑n((k+1)3−k3)(n+1)3−1(n+1)3−1k=1∑nk2====k=1∑n(3k2+3k+1)3k=1∑nk2+3k=1∑nk+k=1∑n13k=1∑nk2+32n(n+1)+n61n(n+1)(2n+1)
Can you find a general pattern to it?
Answer is 301n(n+1)(2n+1)(3n2+3n−1). You can double check your work by induction.
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I appriciate what you did and Thank you for the same.
I have a question- Will this pattern work to find general sum of any power of n?
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Study Bernoulli's Numbers. You'll get the answer for nth power sums. :)
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For positive integer powers, yes.
Study Faulhaber's Formula for a General Idea. Also visit Bernoulli's numbers if it interests you. :)
1/30(n+1)(2n+1)(3n^2+3n-1)
(6x + 15x + 10x - x) / 30
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Even I know the formula but how to derive it is much more important.
Hm, that simplifies to just x. Are you missing some exponents?