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## #1 2017-01-09 05:57:13

Eulero
Member
Registered: 2016-09-14
Posts: 9

### Partial sum formula of a series by recursion

Hi,
I haven't been here for a while, but now i'm back with something new. I found a formula that give the result of the partial sum of the series:
$\sum_{n=1}^{m} n^k$
For each k positive integer. With recursion i mean: do you want the partial sum formula for n=3? You need to know the partial sum formula for n=2 and for that you need partial sum formula for n=1;etc.
This is the formula:

$\LARGE \sum_{n=1}^{m} n^k = \frac{m^{k+1}+km^{k}- \sum\limits_{n=0}^{m-1} n^{k} \left [ n \left [ \left ( 1+\frac{1}{n} \right ) ^k-1 \right ] -k \right ]}{k+1}$

It works perfectly!
Before I publish the proof i really would like your judge:is it a useful formula? Or it's less interesting than i think?
I thank you for every answer.

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## #2 2017-01-09 17:22:49

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,408

### Re: Partial sum formula of a series by recursion

Hi;

That does indeed work for what I have tested but there is already a simple closed form for the first sum.

Does your solution have something special that makes it better for something? If so, what?

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #3 2017-01-09 23:06:48

Eulero
Member
Registered: 2016-09-14
Posts: 9

### Re: Partial sum formula of a series by recursion

Well, I never saw the question by this perspective. I think that my formula is simpler because doesn't need the knowledge of bernoulli numbers that are really hard to remember. And if k gets really high my formula is long because of the recursion but still practicable: i challenge everyone at remembering for a long time a big part of Bernoulli sequence. So should i post the proof or it's a naive work?

Last edited by Eulero (2017-01-09 23:07:38)

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## #4 2017-01-10 00:47:01

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,408

### Re: Partial sum formula of a series by recursion

Hi;

I would say go ahead and give it a shot. Where do you intend to publish?

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #5 2017-01-10 01:32:08

Eulero
Member
Registered: 2016-09-14
Posts: 9

### Re: Partial sum formula of a series by recursion

I think that i used the wrong word (I'm italian), with publish i mean post here. However this is my proof:

Knowing that:
(excuse me the under subscript is 0)
We can rewrite the series so:

Now i notice that delta_n appears m-n times in the series:

We must work a little more for our formula:

QED

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## #6 2017-01-10 01:40:32

Member
From: Planet Mars
Registered: 2016-11-15
Posts: 155

### Re: Partial sum formula of a series by recursion

Hi, Eulero
The formula is too complicated to remember. It can be remembered much more easily by Implementing Faulhaber's formula. It says,
1ˣ +2ˣ +3ˣ +4ˣ ......nˣ =[1/(1+x)](aB₀nˣ⁺¹+bB₁nˣ+cB₂nˣ⁻¹............yBₓn), where Bₙ is the nth Bernoulli no. and a,b,c,.....y are the consecutive terms of (x+1)th row of Pascal's Triangle.
For e.g.
1¹¹+2¹¹+3¹¹+4¹¹.....7¹¹=(1/12)(1B₀n¹²+12B₁n¹¹+66B₂n¹⁰+220B₃n⁹+495B₄n⁸+792B₅n⁷+924B₆n⁶+792B₇n⁵+495B₈n⁴+220B₉n³+66B₁₀n²+12B₁₁n)
(1+B)ⁿ⁺¹-Bₙ₊₁=0
For e.g.
for,B₁ n=1 so
(1+B)²-B₂=0
⇒1+B₂+2B-B₂=0
⇒1=2B=0
⇒B=-0.5
However in this ( the formula for sums of powers )formula B₂=+0.5

Last edited by iamaditya (2017-01-10 01:41:52)

Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

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