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This question already has an answer here:

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?

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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.

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

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

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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.

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