Question

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

(ii) the first n positive integers

Answer 1

(i)

Let S = 1 + 2 + 3 + . . . + 1000

Using the formula $S_n=\left(\frac{n}{2}\right)[a+l]$; for the sum of the first n terms of an AP, we have

$S_{1000}=\frac{1000}{2}(1+1000)$

$=500\times 1001$

$=500500$

So, the sum of the first 1000 positive integers is 500500.

(ii)

Let $S_n$ = 1 + 2 + 3 + . . . + n

Here a = 1 and the last term l is n.

Therefore, $S_n=\left(\frac{n}{2}\right)[1+n]$

So, the sum of first n positive integers is given by

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