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

student34

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

Homework Helper

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

Homework Helper
So would this work too?

2n+1
Ʃ (k) = (2n+1)((2n+1)+1)/2
k=1

Sure. Same thing.

Homework Helper
Dearly Missed

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