# General sum

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
4 years, 3 months ago

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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.

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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.

- 4 years, 2 months ago

Well, it's true. I was just giving a general idea! :)

- 4 years, 2 months ago

For positive integer powers, yes.

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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

Staff - 4 years, 2 months ago