What is the solution to the unsolved equation?

[kazimo]

Hey,
I am kinda new here but here's 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 isn't any other obvious answer. ( These were the first numbers I tried when I began trying to solve this problem.

Kazim

 

[rajesh]

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

Is this true?

 

[Gokul43201]

Clearly, it is not.

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

 

[rhia]

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

Is this true?

It was a typo ...it should have been

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

 

[kazimo]

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

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

I will be careful in the future

 

[dttubbs]

I believe 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.

 

[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

 

[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.