I'm writing up my maths notes to the computer, while looking over maths notes for series I spotted another way of proving the sum of natural numbers to n (to what I've been taught).(adsbygoogle = window.adsbygoogle || []).push({});

By definition:

[tex]\sum_{k=1}^n k = n + (n-1) + (n-2) + \ldots + (n - [n-2]) + (n - [n-1])[/tex]

Separating terms we get:

[tex]\sum_{k=1}^n k = (n + n + \ldots + n) - [1 + 2 + 3 + \ldots + (n-2) + (n-1)][/tex]

[tex]\sum_{k=1}^n k = n^2 - \sum_{k=1}^{n-1} k[/tex]

[tex]\sum_{k=1}^n k = n^2 - \left( \left[ \sum_{k=1}^{n} k \right] - n \right)[/tex]

[tex]2\sum_{k=1}^n k = n^2 + n[/tex]

[tex]\sum_{k=1}^n k = \frac{n(n+1)}{2}[/tex]

So is this sufficient proof please?

Furthermore I was trying to prove from first principles that:

[tex]\frac{d}{dx} \left( x^n \right) = nx^{n-1}[/tex]

For all real values of n, but so far I've only been able to prove it for all positive integers using binomial theorem. Could someone give me a clue at least please?

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# Couple of Questions on proof

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