Question

What is the sum of the first 1000 positive integers?

Answer 1

See, the formula says:

\[1+2+3....................+n = n(n+1)/2\]

Here, n = 1000

Put in the formula and calculate the sum then...

Answer 2

\[\frac{1000 \times 1001}{2} = 500 \times 1001\]

Multiply this and get the answer...

Answer 3

\[S_n=\frac{n}{2}[2a+(n-1)d]\]OR\[S_n=(a+l) \times \frac{n}{2}.\]

Answer 4

THere was a formula for that? I never knew that. No wonder I was getting a wrong answer. Well, now that I think about it, it might just be me being tired...It makes perfect sense now.

Answer 5

500,500

Answer 6

Yeah, 500*1001 = 500,500..

Answer 7

FYI: It's 6:00 where I am now.

Answer 8

There are three values I want to share with you:

1. Sum of First n natural Numbers:
\[1+2+3+4+5...............+n = \frac{n(n+1)}{2}\]

2. Sum of squares of First n natural Numbers:
\[1^2+2^2+3^2+4^2+5^2...............+n^2 = \frac{n(n+1)(2n+1)}{6}\]

3. Sum of Cubes of First n natural Numbers:
\[1^3+2^3+3^3+4^3+5^3...............+n^3 = \frac{n^2(n+1)^2}{4} = [\frac{n(n+1)}{2}]^2\]

Answer 9

This is the formula for a sequence which must be in Arithmetic Progression..

If value of d the common difference is same you only can then use this formula..

And in case of squares and cubes you cannot use this what so called Universal Formula @maheshmeghwal9 ..

Answer 10

oops sorry i forgt that:)
but is that cube or squre GP?

Answer 11

@waterineyes

Answer 12

No, it is not that..
But it is not the universal formula my dear friend..

Answer 13

k i must take back my words.
thanx for capturing my fault.

Answer 14

Welcome buddy..