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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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## Comments

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TopNewestStudy Faulhaber's Formula for a General Idea. Also visit Bernoulli's numbers if it interests you. :)

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1/30(n+1)(2n+1)(3n^2+3n-1)

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Hint:$\begin{aligned} \sum_{k=1}^n \left( (k+1)^2 - k^2 \right) &= &2 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\ \displaystyle (n+1)^2 - 1 &=& 2 \color{#3D99F6}{\sum_{k=1}^n k} + n \\ \displaystyle \sum_{k=1}^n k &=& \frac{n(n+1)}2 \end{aligned}$

And

$\begin{aligned} \sum_{k=1}^n \left( (k+1)^3 - k^3 \right) &= & \sum_{k=1}^n (3k^2 + 3k + 1) \\ \displaystyle (n+1)^3 - 1 &=& 3\color{#3D99F6}{ \sum_{k=1}^n k^2} +3 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\ \displaystyle (n+1)^3 - 1 &=& 3\color{#3D99F6}{ \sum_{k=1}^n k^2} +3\frac{n(n+1)}2 + n \\ \sum_{k=1}^n k^2 &=& \frac16 n(n+1)(2n+1) \\ \end{aligned}$

Can you find a general pattern to it?

Answer is $\frac1{30} n(n+1)(2n+1)(3n^2 + 3n-1)$. You can double check your work by induction.

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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.

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I appriciate what you did and Thank you for the same.

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(6x + 15x + 10x - x) / 30

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Hm, that simplifies to just $x$. Are you missing some exponents?

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Even I know the formula but how to derive it is much more important.

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