2305This question also has a geometric approach to it:
Let the length of the innermost square be 1.
Area of the 4 innermost squares = 4 *(1x1) = 4 * 13
Area of the 8 outer squares = 8 *(2x2) = 4 * 23
Area of the 12 outer squares = 12 *(3x3) = 4 * 33
Length of each side = 2(1+2+3)
Therefore,
[2(1+2+3)]2 = 4 (13+23+33)
or,
(1+2+3)2 = 13+23+33
This can be generalized to n terms,
13+23+33.....+n3 = (1+2+3+...+n)2 = (n(n+1)2)2
62I guess I had a similar one... on sum of squares..
The same logic works
A combinatorial proof...
Number of ways to chose 4 numbers w, x, y, z from {1, 2, ... n+1} so that z is greater than each of the other three
The number of ways by double counting...
when z=2 no of ways = 13 (w, x and y can be chosen in 1 way each)
when z=3 no of ways = 23 (w, x and y can be chosedn in 2 ways each)
when z=4 no of ways = 33 (w, x and y can be chosedn in 3 ways each)
...
...
...
when z=n+1 no of ways = n3 (w, x and y can be chosedn in n ways each)
thus one counting gives us
13+23+.... n3
NOw explain the other way .. I must leave somethign for you guys to think.. [3]
11The proof that I had in mind while I gave this one was different Nishant sir.....[4]
Anyway nice tech....
62
Lokesh Verma
·2009-10-17 01:43:45
The proof you have is
(1+x)4
(1+1)4 = .....
take the summation and see that a lot of terms of 4th power cancel each other..
(Is that in your mind?)
Bcos I dont know any other "good" proof!
11
Devil
·2009-10-17 08:42:29
Nope!
AAre the proof that I've in mind is not mine original, it's from a book and the proof reveals a few interesting things - that's the reason I came up with that particular thing, else u can manufacture a thousand and one proves of this theorem.....
This is what I want :-
Express k3 as an identity in terms of bionomial coefficients......
62
Lokesh Verma
·2009-10-17 22:13:35
hmm.. I get what you are hinting at...
k3 = 6 kC3 + ... + ....
And then take summation?
62
Lokesh Verma
·2009-10-20 09:26:35
k3 =6xkC3 + 2xkC2 + kC1
First check if what i have done is correct (Algebraically and Logically)
Then finish it off....
(I hope this is what soumik had in mind... !!)
62
Lokesh Verma
·2009-10-21 01:15:32
Some one finishing this off?
24Sir was reffering to this one for squares..
http://targetiit.com/iit-jee-forum/posts/squares-1042.html[1]
Soumyabrata Mondal Antonella Cupillari and Warren Lushbaugh's proof Upvote·0· Reply ·2013-03-05 04:57:50