Question

Find the sum of :(i) the first 1000 positive integers (ii) the first n positive integers

Answer 1

i) the formula for finding sum of first n integers is
n× (n+1)÷ 2
here n = 1000
so

1000(1001)/2 = 500500

ii) similarly doing we get n^ 2 + n ÷ 2

Answer 2

Answer:

(i) 500500

(ii) n(n+1)/2

Step-by-step explanation:

(i) The sum of the first 1000 positive integers that we have to calculate.

Hence, we have to find  

1+2+3+4+5+6+ ..... +1000

Now, this is an A.P. series with first term 1, last term 1000 and the number of terms in the series is 1000.

Hence, 1+2+3+4+5+6+ ..... +1000

=1000(1+1000)/2 {Sum of an A.P is given by [Number of terms( first term+ last term)/2]}

=500500. (answer)

(ii) Therefore the sum of the first n positive integers will be n(n+1)/2

{Just replacing 1000 in the previous problem with n} (Answer)