What is the sum of first 1000 whole numbers
let be, 1+2+3+………1000
∴Sn=………
S 1000 =
2
(1+1000)
=500 X 1001
=………..
∴The sum of first 1000 numbers =
Answers
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.
Answer:
the answer is 500500
Step-by-step explanation:
1,2,3...999,1000 = 500500