I have doubt solve a question the question is first n natural numbers. This questions is present in my book in exercise 15.2.

With Regards
Shivam Tiwari

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Solution

$S_n=\frac{n(n+1)}{2}$
Proof:

$1+2+...+(n-1)+n=S_n$

$(1+2+....+(n-1)+n) + (1+2+....+(n-1)+n) = 2S_n$

(1+2+....+(n-1)+n) + (n+(n-1)+....+2+1) = $2S_n$

if we look at the sum above we notice that we have n columns of numbers and that on each column we have two numbers with the sum n+1, so:

$n(n+1) = 2S_n$

$S_n=\frac{n(n+1)}{2}$

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I have doubt solve a question the question is first n natural numbers. This questions is present in my book in exercise 15.2. With Regards Shivam Tiwari

Proof: (1+2+....+(n-1)+n) + (n+(n-1)+....+2+1) = if we look at the sum above we notice that we have n columns of numbers and that on each column we have two numbers with the sum n+1, so:

I have doubt solve a question the question is first n natural numbers. This questions is present in my book in exercise 15.2.

With Regards
Shivam Tiwari