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y2o2Best ResponseYou've already chosen the best response.0
\[\large {^{2n} C _n} = {(^n C _0)}^2 +{(^n C _1)}^2 +{(^n C _2)}^2 +...+{(^n C _n)}^2 \]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
So in summation notation, show that\[\binom{2n}{n}=\sum_{i=0}^n \binom{n}{i}^2\]
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
you can use induction..
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
Use Vandermonde's identity
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
well have you tried it?
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
since it was from 2 months ago..
 one year ago

y2o2Best ResponseYou've already chosen the best response.0
Yes , and it came up with nothing
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
\[LHS: \binom{2n}{n} x^{n} \] \[RHS\left[\binom{n}{0}\binom{n}{n}+\binom{n}{1}\binom{n}{n1}+\binom{n}{2}\binom{n}{n2}+....+\binom{n}{n2}\binom{n}{2}+\binom{n}{n1}\binom{n}{1}+\binom{n}{n}\binom{n}{0}\right] *x^n\] \[By ~Symmetry: \binom{n}{n} = \binom{n}{0} ; \binom{n}{n1} = \binom{n}{1} \] \[hence~ RHS = \left[\binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2}\right] x^n\] \[hence, coeeficients~ of~ x^n~ are~ the ~same \] \[\therefore, \binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2} = \binom{2n}{n} \]
 one year ago

vamgaduBest ResponseYou've already chosen the best response.0
use this. its better. http://upload.wikimedia.org/wikipedia/en/math/4/0/6/406c9e30b2566d50df58a5dc3315d1d2.png
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
whats wrong with the method i used?
 one year ago

vamgaduBest ResponseYou've already chosen the best response.0
nothing. I said its easier to use vandermonde's identity.
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
okay; well i dont know how to use that identity lol
 one year ago

vamgaduBest ResponseYou've already chosen the best response.0
just plug in n every where except at k(its a summation variable).
 one year ago

y2o2Best ResponseYou've already chosen the best response.0
thank you, i got it now :)
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
Sorry I made a typo at the end: \[\therefore, \binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2} = \binom{2n}{n} ^2\]
 one year ago

y2o2Best ResponseYou've already chosen the best response.0
yeah , i realized that :)
 one year ago
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