Linked Questions

1
vote
2answers
2k views

The sum of the fourth powers of the first $n$ positive integers [duplicate]

I am studying mathematical induction and most of the times I have to prove something. Like, for example: $1 + 4 + 9 + ...+ n^2 = \frac{n(n+1)(2n+1)}{6}$ This time I found a question that ask me to ...
3
votes
2answers
114 views

How to prove that $1^r + 2^r +… + n^r = a_1 n^1 + a_2 n^2 +… + a_{r+1} n^{r+1}$? [duplicate]

I see this equation $$ 1^r + 2^r +... + n^r = a_1 n^1 + a_2 n^2 +... + a_{r+1} n^{r+1} $$ in Introduction to Linear Algebra, where $a_k$s are some constants. How can I prove it?
5
votes
0answers
231 views

Finding a general way to sum $ \sum_{i=1}^{i=n} i^k$ for a given 'k' using elementary highschool calculus [duplicate]

Today morning I had thought of a wonderful way to calculate the general sum of $$ \sum_{i=0}^{i=n} i^n$$ Using just things taught in elementary high school calculus. So, my method is as follows, First ...
0
votes
2answers
46 views

How to determine the sum of a series that is neither geometric nor arithmetic but quadratic or cubic? [duplicate]

How to determine the formula of the sum of a series given its $n$th-term formula like: $$U_n= n^2+n$$ or $$U_n= 6n^2 -12n + 5$$
0
votes
0answers
42 views

What is the sum of the given special series?any specific formula? [duplicate]

$1^n$+$2^n$+$3^n$+$\dots$+$m^n$. $(m,n\in\mathcal N)$ $m,n\ge2$
127
votes
32answers
68k views

Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$

I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals: $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ I really ...
39
votes
17answers
4k views

I'm looking for some mathematics that will challenge me as a year $12$ student. [closed]

I am an upcoming year $12$ student, school holidays are coming up in a few days and I've realised I'm probably going to be extremely bored. So I'm looking for some suggestions. I want a challenge, ...
12
votes
14answers
575 views

Different ways to come up with $1+2+3+\cdots +n=\frac{n(n+1)}{2}$

I am trying to compile a list of the different ways to come up with the closed form for the sum in the title. So far I have the famous Gauss story, the argument by counting doubletons, and using ...
7
votes
12answers
3k views

Compilation of proofs for the summation of natural squares and cubes

I want to know different proofs for the following formulas, $$ \sum_{i=1}^n{i^2} = \frac{(n)(n+1)(2n+1)}{6} $$ $$ \sum_{i=1}^n{i^3} = \frac{n^2(n+1)^2}{2^2} $$ Please do not mark this as duplicate, ...
23
votes
5answers
17k views

A formula for the power sums: $1^n+2^n+\dotsc +k^n=\,$?

Is there explicit formula for the expression $1^n + 2^n + \dotsc + k^n\,$? I know that for $n=1$ the explicit formula becomes $S=k(k+1)/2$ and for $n=3$ the formula becomes $S^2$. But what about ...
14
votes
2answers
4k views

Finite Sum of Power?

Can someone tell me how to get a closed form for $$\sum_{k=1}^n k^p$$ For $p = 1$, it's just the classic $\frac{n(n+1)}2$. What is it for $p > 1$?
9
votes
5answers
1k views

How Are the Solutions for Finite Sums of Natural Numbers Derived?

So, I've been learning set theory on my own (Lin, Shwu-Yeng T., and You-Feng Lin. Set Theory: An Intuitive Approach. Houghton Mifflin Co., 1974.) and have come across infinite sums of natural numbers. ...
9
votes
3answers
1k views

Closed form for $1^k + … + n^k$ (generalized Harmonic number)

This question must have been asked, it's just very hard to search for such questions. I'm looking for the cleanest method I can find for getting a closed form formula for $\sum_{i=1}^n i^k$ ...
9
votes
5answers
4k views

How is Faulhaber's formula derived?

I have been wanting to understand how to find the sum of this series. $$1^p + 2^p + 3^p +{\dots} + n^p$$ I am familiar with Gauss' diagonalised adding trick for the sum of the first $n$ natural ...
8
votes
3answers
2k views

Formula for $1^k+2^k+3^k…n^k$ for $n,k \in \mathbb{N}$

So I've been looking for a formula where I can input the parameter $k$ and it will give me a formula for $1^k+2^k+3^k...+ n^k$ with $n,k \in \mathbb{N}$. The result is always a polynomial with $k+1$ ...

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