Question

prove that :

Answer 1

\[\large {^{2n} C _n} = {(^n C _0)}^2 +{(^n C _1)}^2 +{(^n C _2)}^2 +...+{(^n C _n)}^2 \]

Answer 2

So in summation notation, show that\[\binom{2n}{n}=\sum_{i=0}^n \binom{n}{i}^2\]

Answer 3

you can use induction..

Answer 4

Use Vandermonde's identity

Answer 5

well have you tried it?

Answer 6

since it was from 2 months ago..

Answer 7

Yes , and it came up with nothing

Answer 8

no it works.

Answer 9

\[LHS: \binom{2n}{n} x^{n} \]
\[RHS\left[\binom{n}{0}\binom{n}{n}+\binom{n}{1}\binom{n}{n-1}+\binom{n}{2}\binom{n}{n-2}+....+\binom{n}{n-2}\binom{n}{2}+\binom{n}{n-1}\binom{n}{1}+\binom{n}{n}\binom{n}{0}\right]
*x^n\]
\[By ~Symmetry: \binom{n}{n} = \binom{n}{0} ; \binom{n}{n-1} = \binom{n}{1} \]
\[hence~ RHS = \left[\binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2}\right] x^n\]
\[hence, coeeficients~ of~ x^n~ are~ the ~same \]
\[\therefore, \binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2} = \binom{2n}{n} \]

Answer 10

use this. its better.
http://upload.wikimedia.org/wikipedia/en/math/4/0/6/406c9e30b2566d50df58a5dc3315d1d2.png

Answer 11

whats wrong with the method i used?

Answer 12

nothing. I said its easier to use vandermonde's identity.

Answer 13

okay; well i dont know how to use that identity lol

Answer 14

just plug in n every where except at k(its a summation variable).

Answer 15

thank you, i got it now :)

Answer 16

You're welcome (:

Answer 17

Sorry I made a typo at the end:
\[\therefore, \binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2} = \binom{2n}{n} ^2\]

Answer 18

yeah , i realized that :)