Diwali specials.....

Remember the formula for the sum of cubes of 1st n naturals.....
\frac{n^2(n+1)^2}{4}
Remember the formula we learnt in progressions for deriving this?

Now derive this using bionomial theorem.....

10 Answers

2305
Shaswata Roy ·

This 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

62
Lokesh Verma ·

I 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]

11
Devil ·

The proof that I had in mind while I gave this one was different Nishant sir.....[4]
Anyway nice tech....

62
Lokesh Verma ·

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 ·

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 ·

hmm.. I get what you are hinting at...

k3 = 6 kC3 + ... + ....

And then take summation?

62
Lokesh Verma ·

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... !!)

11
Devil ·

Yup!

62
Lokesh Verma ·

Some one finishing this off?

24
eureka123 ·

Sir was reffering to this one for squares..
http://targetiit.com/iit-jee-forum/posts/squares-1042.html[1]

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