Question

sum of even natural numbers between 1to1000​

Answer 1

Answer:

The sum of first n natural numbers is given by the formula: n(n+1)/2.

This can be derived as follows:

Let's assume there's an Arithmetic Progression (An Arithmetic Progression is a series of numbers which starts with some number and each successive number is greater than the number before it by a common difference of d) whose first term is a and the common difference is d, so the terms in the AP are: a, a+d, a+2d, a+3d and so on. It can easily be seen that the nth term in the AP ca be written as: a+(n-1)d. Let S represent the sum of first n terms of the AP. So,

S=a + (a+d) + (a+2d) + . . . + (a+(n-1)d) and also

S=(a+(n-1)d) + (a+(n-2)d) + . . . + (a+d) + a

(In the 2nd equation, the terms to be summed are just written in the reverse order).

Adding the 2 equations gives us:

2S= 2a+(n-1)d + 2a+(n-1)d + 2a+(n-1)d + . . . n times

=>2S= n(2a+(n-1)d)

=>S= n(a+(n-1)d/2)

And as we can observe that the set of natural numbers is an AP with a=1 and d=1. Putting these in S,

S= n(1+n/2–1/2)

=>S= n(n/2+1/2)

=>S= n(n+1)/2

So, the sum of numbers from 1 to 1000=1000*1001/2=500500.