approximation of sum of integers to the power of 4

Also, I know how to prove sum of squares but how would you do power of 4?

Thanks!

Also, I know how to prove sum of squares but how would you do power of 4?

Thanks!

Can you expand on what you're trying to do?

Also, this makes no sense.

Also, I know how to prove sum of squares but how would you do power of 4?

Thanks!

In suspect s(he) means 1^4 + 2^4 + 3^4 + 4^4 + .... + n^4

How closely do you want to approximate it? Zero is a pretty good approximation, only out by at most 10^20.

Otherwise, the sum is going to be given by a fifth-power expression, and will be divisible by $\frac{1}{6}n(n+1)(2n+1)$ being the corresponding sum of squares; hence it is a quadratic times that thing. (It's a fact, I seem to remember, that the sum of nth powers is a multiple of the sum of n-2th powers.)

That thing is 338350, so we're looking at roughly $100^2 \times 338350 = 3,383,500,000$. I'll knock some sig figs off that because there's no reason at all for the quadratic to have leading coefficient 1; let's go with $1 \times 10^{10}$ as the possibly-correct order of magnitude.

Alternatively:

$\sum_{i=1}^n i^4 = n \mathbb{E}(X^4)$ where $X$ follows the uniform distribution on $\{1, 2, \dots, n \}$. Also $\sum_{i=2}^{n+1} i^4 = n \mathbb{E}((X+1)^4)$, so $(n+1)^4-1 = n \mathbb{E}((X+1)^4 - X^4) = n \mathbb{E}[4X^3+6X^2+4X+1]$.

That is, $n^3+4n^2+6n+4 = 4 \mathbb{E}(X^3) + 6 \mathbb{E}(X^2) + 4 \mathbb{E}(X) + 1$. $\mathbb{E}(X) = \frac{n+1}{2}$; $\mathbb{E}(X^2) = \frac{1}{2} (n+1)(2n+1)$ by the usual formula for the sum of squares; hence $\mathbb{E}(X^3)$ can be calculated by simply rearranging.

if you consider $\displaystyle\sum_{r=1}^{n} (r+1)^k - r^k$ you should be able to derive a formula for general integer values of k in terms of the sums of the previous powers

Yes that is the way I figured out how to do! But I was asked to approximate it in my practise interview and couln't think of any way to do it without using the method above.

Yes that is the way I figured out how to do! But I was asked to approximate it in my practise interview and couln't think of any way to do it without using the method above.

sum of n is O(n²)
sum of n² is O(n³)
sum of n³ is O(n⁴)
sum of n⁴ is O(n⁵)

therefore 100⁵

bit crude but there are approximations and approximations

But how did you get to that approximation of putting a pwoer up?

look at the formulas of

Sum(n)
Sum(n²)
Sum(n³)

ohhhh!!! I feel so stupid! I got it now! thank you so much

Could you possibly explain to me about:

will be divisible by being the corresponding sum of squares; hence it is a quadratic times that thing. (It's a fact, I seem to remember, that the sum of nth powers is a multiple of the sum of n-2th powers.)

Sorry I don't really understand

Otherwise, the sum is going to be given by a fifth-power expression, and will be divisible by being the corresponding sum of squares; hence it is a quadratic times that thing. (It's a fact, I seem to remember, that the sum of nth powers is a multiple of the sum of n-2th powers.)