# Mathematical Proof By Induction of $1 + 2 +\dots + n = (n + 1)n/2$ [duplicate]

Trying to prove that $$1 + 2 +\dots + n = (n + 1)n/2$$.

I have let $$n = 1$$ in the basis step, which lead to $$1=1$$, so it is true for $$n = 1$$.

For the induction step, I have assumed that $$k\geq 1$$ and let $$n = k$$ leading to:

$$1 + 2 + . . . + k = (k + 1)k/2$$

For which we then have to use to prove that the statement is valid for $$n = k + 1$$, which is where I get confused because this the statement is supposed to become:

$$1 + 2 + . . . + k + (k + 1) = (k + 2)(k + 1)/2$$

While I thought it would have just been

$$1 + 2 + . . . + (k + 1) = (k + 2)(k + 1)/2$$

Via the substitution of $$k = k + 1$$. Why is there still a singular $$k$$ in the $$k + 1$$ statement?

## marked as duplicate by N. F. Taussig, user10354138, Don Thousand, Nicolas FRANCOIS, Parcly TaxelNov 4 '18 at 14:15

• $1+2+\dots +(k+1)$ is exactly the same as $1+2+\dots +k+(k+1)$. – lulu Nov 3 '18 at 16:04
• – Lord Shark the Unknown Nov 3 '18 at 16:05

1 + 2 + . . . + k + (k + 1)

and

1 + 2 + . . . + (k + 1)

are exactly the same: in both cases it is the sum of all numbers from $$1$$ to $$k+1$$ ... which includes $$k$$. It's just that the first representation makes the fact that this includes $$k$$ explicit, while the second does not.

• Thanks, this is actually quite simple – Ben Beaumont Nov 3 '18 at 16:07
• @BenBeaumont You're welcome! :) – Bram28 Nov 3 '18 at 16:08

Basic algebra is what's causing the problems: you reached the point

$$\frac{1}{2}K\color{red}{(K+1)}+\color{red}{(K+1)}\;\;\;\:(**)$$

Now just factor out the red terms:

$$(**)\;\;\;=\color{red}{(K+1)}\left(\frac{1}{2}K+1\right)=\color{red}{(K+1)}\left(\frac{K+2}{2}\right)=\frac{1}{2}(K+1)(K+2)$$

It is because we assume that the statement is true for 'K' and 'K+1'.We have checked that the statement is true for k=1. Therefore, by what we have just seen, it must be true for k=2 and hence for the next integer 3. proceeding Inductively, we see that formula has to be true for all integers.