Thursday 19 March 2015

algebra precalculus - Sum of first n natural numbers proof

I know how to prove this by induction but the text I'm following shows another way to prove it and I guess this way is used again in the future. I'm confused by it.




So the expression for first n numbers is:
$$\frac{n(n+1)}{2}$$



And this second proof starts out like this. It says since:



$$(n+1)^2-n^2=2n+1$$



Absolutely no idea where this expression came from, doesn't explain where it came from either.



Then it proceeds to say:

\begin{align}
2^2-1^2&=2*1+1 \\
3^2-2^2&=2*2+1\\
&\dots\\
n^2-(n-1)^2&=2(n-1)+1\\
(n+1)^2-n^2&=2n+1
\end{align}
At this point I'm completely lost.
But it continues to say "adding and noting the cancellations on the left, we get"
\begin{align}

(n+1)^2-1&=2(1+2+...+n)+n \\
n^2+n&=2(1+2+...+n) \\
(n(n+1))/2&=1+2+...+n
\end{align}



Which proves it but I have no clue what has happened. I am entirely new to these math proofs. Im completely lost. I was great at high school math and calculus but now I haven't got the slightest clue of what's going on. Thanks

No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...