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Can anyone tell me how to calculate general sum of sequence till \(n\) . \[1^{4}+2^{4}+3^{4}+ \ldots +n^{4}\]

Note by Akshat Sharda 2 years, 8 months ago

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2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

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Hint:

\[ \begin{eqnarray} \sum_{k=1}^n \left( (k+1)^2 - k^2 \right) &= &2 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\ \displaystyle (n+1)^2 - 1 &=& 2 \color{blue}{\sum_{k=1}^n k} + n \\ \displaystyle \sum_{k=1}^n k &=& \frac{n(n+1)}2 \end{eqnarray} \]

And

\[ \begin{eqnarray} \sum_{k=1}^n \left( (k+1)^3 - k^3 \right) &= & \sum_{k=1}^n (3k^2 + 3k + 1) \\ \displaystyle (n+1)^3 - 1 &=& 3\color{blue}{ \sum_{k=1}^n k^2} +3 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\ \displaystyle (n+1)^3 - 1 &=& 3\color{blue}{ \sum_{k=1}^n k^2} +3\frac{n(n+1)}2 + n \\ \sum_{k=1}^n k^2 &=& \frac16 n(n+1)(2n+1) \\ \end{eqnarray} \]

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 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 \(n\)?

Study Bernoulli's Numbers. You'll get the answer for nth power sums. :)

@Satyajit Mohanty – 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.

@Pi Han Goh – Well, it's true. I was just giving a general idea! :)

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

Hm, that simplifies to just \(x\). Are you missing some exponents?

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

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`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestHint:\[ \begin{eqnarray} \sum_{k=1}^n \left( (k+1)^2 - k^2 \right) &= &2 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\ \displaystyle (n+1)^2 - 1 &=& 2 \color{blue}{\sum_{k=1}^n k} + n \\ \displaystyle \sum_{k=1}^n k &=& \frac{n(n+1)}2 \end{eqnarray} \]

And

\[ \begin{eqnarray} \sum_{k=1}^n \left( (k+1)^3 - k^3 \right) &= & \sum_{k=1}^n (3k^2 + 3k + 1) \\ \displaystyle (n+1)^3 - 1 &=& 3\color{blue}{ \sum_{k=1}^n k^2} +3 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\ \displaystyle (n+1)^3 - 1 &=& 3\color{blue}{ \sum_{k=1}^n k^2} +3\frac{n(n+1)}2 + n \\ \sum_{k=1}^n k^2 &=& \frac16 n(n+1)(2n+1) \\ \end{eqnarray} \]

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 appriciate what you did and Thank you for the same.

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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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Study 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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(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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