Question

the sum of first n off numbers is ​

Answer 1

Answer:

Sum of the First n Natural Numbers. We prove the formula 1+ 2+ ... + n = n(n+1) / 2, for n a natural number. There is a simple applet showing the essence of the inductive proof of this result

Step-by-step explanation:

Sum of the First n Natural Numbers. We prove the formulaSum of the First n Natural Numbers. We prove the formula 1+ 2+ ... + n = n(n+1) / 2, for n a natural number. There is a simple applet showing the essence of the inductive proof of this result, for n a natural number. There is a simple applet showing the essence of the inductive proof of this result

Answer 2

Answer:

Your question seems to be unclear, but I will answer it anyway.

Step-by-step explanation:

Sum of first n numbers of an AP is given by the formula

$S_n=\frac{n}{2}[2a+(n-1)d]$

where a is the starting term of the AP and d is the common difference between the terms.

The set of natural numbers form an AP as follow:

1, 2, 3, 4, 5.......

where a=1 and d=1

Plugging in the values of a and d into the formula, the sum of the first n natural numbers is

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

So, the sum of the first n natural numbers is the product of n, the successor of n and 1/2.

Hope it helps!

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