I din know under which way to post this...
but this was a brilliant brilliant method to prove that sum of squares first n natural numbers is given by permutation and counting principle..
I am giving the hint here and I hope that you guys Post the complete solution...
First Statement:
Count the number of ways in which you can chose 3 numbers x, y, z (with repetition) from the set {1, 2, 3, .... n+1} such that z is the largest.. and z > x and z>y
Try and figure out or complete the proof
(For all users except Prophet Sir and Kaymant Sir) [1]
After we have done this, we will also try and list the other conventional and if possible some more non conventional methods to prove the sum :)
62No one! atleast give a few steps.. I will keep helping u on the way...
62Reading this proof inspired me to add these on wikipedia ;)
Btw does this have a Hungarian history!??
341i happened upon it in a book recently. i remember some association with Polya. can't be sure though
1as u hv asked about finding the proof for first natural nos., here is the proof:
we have n numbers when we take z to be the greatest repeatedly out of x,y,z from the set{1,2,....n+1}.there are n+1 ways after all repeatations.
after each repeatation 2 nos. are left ,i.e, (n-1)ways.
therefore
(n+1)C(n-1) = n(n+1)/2.
proved.
reply soon please............!!!!!!!!!!!!
3sir wat i got is a summation but that wont help here
here it is no of ways \sum_{2}^{n+1}{r(r-1)^{2}}
62no it is not that..
see more closely..
when z goes from 2 to n+1 .. z can be chosen in one way.. because you have fixed z.
62three is no reason why.
i gave this proof because it adds a new perspective to the whole thing :P