Special Series

[simonlai90]

How to solve ?(x[sup:29fea03j]n[/sup:29fea03j]) from x=1 to n-1 ? what is the value in term of x and n? Thx... :wink:

 

[galactus]

Are you sure you don't mean something like:

\(\displaystyle \sum_{k=1}^{n-1}x^{k}=\frac{x^{n}-x}{x-1}\)?.

Your problem says \(\displaystyle \sum_{x=1}^{n-1}x^{n}\)

 

[simonlai90]

My problem is not on geometric series.

I can make it more clear,

{ ?(x[sup:2raytv1h]n[/sup:2raytv1h]) from x=1 to n-1 } = 1[sup:2raytv1h]n[/sup:2raytv1h] + 2[sup:2raytv1h]n[/sup:2raytv1h] + 3[sup:2raytv1h]n[/sup:2raytv1h] + 4[sup:2raytv1h]n[/sup:2raytv1h] + ... + (n-1)[sup:2raytv1h]n[/sup:2raytv1h]

so what is the value of sum of the equation in term of x and n?

 

[soroban]

Hello, simonlai90!

\(\displaystyle \text{How to solve: }\:\sum^n_{k=1} k^r\)

\(\displaystyle \text{What is the value in term of }n?\)

This is not an elementary problem.


I can give you the first few formulas . . .

\(\displaystyle \sum^n_{k=1} k \;\;=\;\tfrac{1}{2}n(n+1)\)

\(\displaystyle \sum^n_{k=1}k^2 \;=\;\tfrac{1}{6}n(n+1)(2n+1)\)

\(\displaystyle \sum^n_{k=1}k^3 \;=\;\tfrac{1}{4}n^2(n+1)^2\)

\(\displaystyle \sum^n_{k=1}k^4 \;=\;\tfrac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)\)

\(\displaystyle \sum^n_{k=1}k^5 \;=\;\tfrac{1}{12}n^2(n+1)^2(2n^2+2n-1)\)

 

[galactus]

Yes, sorry about that. I was just making sure.

Besides what Soroban listed, the general formula is kind of complicated and involves Bernouilli numbers.

See here:

http://www.math.rutgers.edu/~erowland/sumsofpowers.html