$1+2+3+...+n-1$ sum of the first $n-1$ terms

Question

I know that $$\sum_{k=1}^{n-1}{k}=\frac{n(n-1)}{2}$$

But if I write $$S = 1+2+3+\ldots+(n-3)+(n-2)+(n-1)$$ and write $S$ backwards, $$S=(n-1)+(n-2)+(n-3)+\ldots+3+2+1$$ Adding two, $$2S = n+n+\ldots+n $$which is adding n ntimes, $$2S=n^2$$ Thus $$S=\frac{n^2}{2}$$

I think there is a mistake, can you help me find the mistake?

Answer 1

You have made a mistake in your calculation. There are $(n - 1)$ terms in your equation. $$S = 1 + 2 + 3 + ..... (n - 2) + (n - 1)$$ $$S = (n - 1) + (n - 2) + .......3 + 2 + 1$$ Adding both, $$2S = n + n + n + n + n....(upto (n - 1) terms$$ $$\therefore 2S = n(n - 1)$$ $$\therefore S = \frac{n(n - 1)}{2}$$ This is your final answer. Hope you like it.

Answer 2

$n+n+n...$

"...which is adding $n$ times" You're adding it for $n-1$ times!!