The method used by Newton is already shown above.
Try to find the sum
S=1+2+3..........+100
by this trick below
S = 1 + 2 + 3 + 4 + ........... 100
S =100+ 99 + 98 + 97 + ........... 1
Add both and see what happens
Note: This is only for class IX  X guys.. Not for every one! So please dont jump in your nose and say it is easy... I strongly think that such tricks should be known to the <11 guys before getting into class XI.

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9 Answers
s=1+2+3+.....+100
s=((1st no+last no)/2)*total numbers
:. s = ((1+100)/2)*100
= 50.5*100
= 5050
Yup.....this trick was given by GAUSS.........
its coool...
but bhaiyya...it was in our text book when i was doin my 9th
hehe
hmm.. it was not in mine :P
I was in ICSE..
So I am giving a level playing ground to everyone.. :P
and yeah once some IXX guy answers this .. i will try to get some mroe complex answers from him...
Carl Freidrich Gauss, known as the 'Prince of Mathematics', solved this sum.
Let the required sum be x./
Therefore, x = 1 + 2 + 3 + ... + 100
Therefore, x = 100 + 99 + 98 + ... + 1
Adding the two equations, we get
2x = (1 + 100) + (2 + 99) + ...(100 + 1)
or, 2x = 101 + 101 + 101 + 101 + 101 + 101 + 101 + 101...(upto 100 times)
[Since there are 2 * 100 or 200 terms, as we are grouping them in pairs of two, number of pairs = 200/2 = 50]
THEREFORE, 2x = 101 * 100
Therefore, x = 101 * 100 2
or, x = 101 8 50
=5050 (ANS.)
Oh! By the way, I study in Class 8, but a teacher in my school taught me this. He came on assignment!
s = 1 + 2 + 3 + .......... + 100
This trick was used by Newton when he was in Std. I or II (maybe)
what he did was, the way Nishant sir wants us to solve this question in!!
s = (100 + 0) + (99 + 1) + (98 + 2) + ............ + (51 + 49) + 50
= 100(50) + 50 = 5000 + 50 = 5050
I hope i m correct...!!