# Unsolved Equation

1. Sep 27, 2004

### kazimo

Hey,
I am kinda new here but heres a problem for you guys:

There is an equation of the form:
((1)^k)+((2)^k)+......+((n-1)^k)+((n)^k) = ((n+1)^k)

This equation is such that all the numbers starting from1 till n are raised to the power of k and added and the result is (n+1)^k. What should n and k be?

Apart from ((1)^2) + ((2)^2) = ((3^2)) There isnt any other obvious answer. ( These were the first numbers I tried when I began trying to solve this problem.

Kazim

2. Sep 27, 2004

### rajesh

((1)^2) + ((2)^2) = ((3^2))?????

Is this true?

3. Sep 27, 2004

### Gokul43201

Staff Emeritus
Clearly, it is not.

But 1+2=3 is true. And there's a more trivial solution : 1^0 = 2^0.

4. Sep 27, 2004

### rhia

It was a typo ...it should have been

(1^1)+(2^1)=(3^1)

5. Sep 27, 2004

### kazimo

I just did a mistake.. it was supposed to be

((1)^1) + ((2)^1) = ((3)^1)

I will be careful in the future

6. Sep 30, 2004

### dttubbs

I belive that if k was an odd number and n was negative then it would be possible to solve it another way. With a negative number in there you can counteract all of the adding of things.

7. Oct 3, 2004

### bomba923

is the original problem like this:

k
sigma (n^c) where c is any real constant
n=1

i was trying to figure that out, but if yours is

k
sigma (c^n) where c is any real constant,
n=1

then i think the sum is {[c^(k+1)]/(c-1)}-[1/(c-1)]

hope this helps

8. Nov 8, 2004

### robert Ihnot

I think about the closest series to that is 1+2+4+8++2^N =2^(n+1)-1. This comes about because $$1 +r +r^2+r^n=\frac{-1+r^_(n+1)}{r-1}$$
but, of course, 2-1=1, so the denominator disappears.

Last edited: Nov 8, 2004