Question

I need some help with Sigma Notation! I'm in Honors PreCalculus and am confused!

Answer 1

Obtain a formula for\[\sum_{k=1}^{n}k^m\]

Answer 2

I seem to be having lots of trouble with Sigma Notation. It's starting to get frustrating because I have 3 problems for my project that I can't quite solve and it's due tomorrow.

Answer 3

Are you sure you typed your question right?

Answer 4

Yes I'm sure. It's k^m and not k^n. That would make it much simpler.

Answer 5

Does the sum start from k=1 or m=1?

Answer 6

k=1

Answer 7

In this case, this is not a easy sum.

Answer 8

If it helps at all, it's given as a tip that \[\sum_{k-1}^{n}k^2 = \frac{ n(n+1)(2n+1) }{ 6}\]

Answer 9

but I don't quite see how that helps

Answer 10

Well it never states m as equal to n+1

Answer 11

Why would that be the case?

Answer 12

Sorry i keep having typos one sec

Answer 13

\[
\sum _{k=1}^n k^3=\frac{1}{4} n^2 (n+1)^2
\]

Answer 14

Ah, so how would that relate to k^m? It's weird since m is not defined in the question

Answer 15

\[
\sum _{k=1}^n k^4=\frac{1}{30} n (n+1) (2 n+1) \left(3 n^2+3 n-1\right)
\]
and so fourth, but there is no easy formula for any m

Answer 16

That makes this really tricky doesn't it

Answer 17

I was able to go as far as to derive that

Answer 18

But I really can't figure out how to do it for the mth power, since the pattern isn't really that easy to determine

Answer 19

For a general m, the sum is defined in terms of waht is called the Harmonic Number

Answer 20

can that be used to generalize a formula here?

Answer 21

We haven't learned about harmonic numbers yet i dont think

Answer 22

I can give you a summary of previous knowledge of this unit?

Answer 23

1) Matrices
2) Geometric Sequences and Series
3) Arithmetic Sequences and Series
4) Pascal's Triangle
5) The Binomial Theorem

Answer 24

http://www.wolframalpha.com/input/?i=sum+k%5Em,+k%3D1+to+k%3Dn

Answer 25

That still uses the Harmonic Number

Answer 26

Is there a way to do it without using that to generalize?

Answer 27

This is the only way. That is why I asked if the summation for k=1, or m=1.
In case, it is m=1, to n. It is a geometric series.

Answer 28

This is a really confusing problem isn't it?

Answer 29

The next one is a bit simpler, maybe if you walk me through it i'll be able to understand how this notation works with exponents?

Answer 30

Find a formula for \[\sum_{k=1}^{n}k^3\]

Answer 31

This summation is done in Calculus for specific m. Like m=1, m=2, m=3 but never for a general m

Answer 32

That is easy, I gave you the answer above

Answer 33

Yes but how is that answer obtained?

Answer 34

See the proof here
http://math.stackexchange.com/questions/320985/how-to-determine-equation-for-sum-k-1n-k3

Answer 35

I don't really need the answers exactly, I need to know how to obtain those answers.

Answer 36

Visit the above link

Answer 37

Will you be here in a few minutes? I wanna read through that and work it out.

Answer 38

There are several approaches. All of them lead to the solution.

Answer 39

What's the link to the solution for the k^4 one?

Answer 40

Ah! I think I got the formula for the first one

Answer 41

Thanks a ton for your help. I believe the answer is \[\sum_{i=1}^{n}k^{m-1}\sum_{k=1}^{n}1\]