New user? Sign up

Existing user? Log in

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 3 years, 4 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

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

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

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.

Log in to reply

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.

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*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

paragraph 2

`[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

Sort by:

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.

Log in to reply

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

Log in to reply

I have a question- Will this pattern work to find general sum of any power of \(n\)?

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

For positive integer powers, yes.

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply