What does 2n on top of the summation expression do diferently than just n?

[student34]

Homework Statement

2n
Ʃ (k)
k=1

The Attempt at a Solution

2n
Ʃ (k) = 2n(2n+1)/2 (This is just a shot in the dark.)
k=1

 

[Dick]

Homework Statement

2n
Ʃ (k)
k=1

The Attempt at a Solution

2n
Ʃ (k) = 2n(2n+1)/2 (This is just a shot in the dark.)
k=1

Sure. If you've shown the sum from 1 to n of k is n*(n+1)/2, then the sum to 2n is just 2n*(2n+1)/2. It's just substitution.

 

[student34]

Sure. If you've shown the sum from 1 to n of k is n*(n+1)/2, then the sum to 2n is just 2n*(2n+1)/2. It's just substitution.

So would this work too?

2n+1
Ʃ (k) = (2n+1)((2n+1)+1)/2
k=1

 

[Dick]

So would this work too?

2n+1
Ʃ (k) = (2n+1)((2n+1)+1)/2
k=1

Sure. Same thing.

 

[Ray Vickson]

Homework Statement

2n
Ʃ (k)
k=1

The Attempt at a Solution

2n
Ʃ (k) = 2n(2n+1)/2 (This is just a shot in the dark.)
k=1

\sum_{k=1}^N k = \frac{N(N+1)}{2},
so if you put N = 2n you get the stated result. If you set N = 2n+1 you get the other result you stated.

RGV