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y2o2 Group TitleBest ResponseYou've already chosen the best response.0
\[\large {^{2n} C _n} = {(^n C _0)}^2 +{(^n C _1)}^2 +{(^n C _2)}^2 +...+{(^n C _n)}^2 \]
 2 years ago

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

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
you can use induction..
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
Use Vandermonde's identity
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.2
well have you tried it?
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.2
since it was from 2 months ago..
 2 years ago

y2o2 Group TitleBest ResponseYou've already chosen the best response.0
Yes , and it came up with nothing
 2 years ago

vamgadu Group TitleBest ResponseYou've already chosen the best response.0
no it works.
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.2
\[LHS: \binom{2n}{n} x^{n} \] \[RHS\left[\binom{n}{0}\binom{n}{n}+\binom{n}{1}\binom{n}{n1}+\binom{n}{2}\binom{n}{n2}+....+\binom{n}{n2}\binom{n}{2}+\binom{n}{n1}\binom{n}{1}+\binom{n}{n}\binom{n}{0}\right] *x^n\] \[By ~Symmetry: \binom{n}{n} = \binom{n}{0} ; \binom{n}{n1} = \binom{n}{1} \] \[hence~ RHS = \left[\binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2}\right] x^n\] \[hence, coeeficients~ of~ x^n~ are~ the ~same \] \[\therefore, \binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2} = \binom{2n}{n} \]
 2 years ago

vamgadu Group TitleBest ResponseYou've already chosen the best response.0
use this. its better. http://upload.wikimedia.org/wikipedia/en/math/4/0/6/406c9e30b2566d50df58a5dc3315d1d2.png
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.2
whats wrong with the method i used?
 2 years ago

vamgadu Group TitleBest ResponseYou've already chosen the best response.0
nothing. I said its easier to use vandermonde's identity.
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.2
okay; well i dont know how to use that identity lol
 2 years ago

vamgadu Group TitleBest ResponseYou've already chosen the best response.0
just plug in n every where except at k(its a summation variable).
 2 years ago

y2o2 Group TitleBest ResponseYou've already chosen the best response.0
thank you, i got it now :)
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.2
You're welcome (:
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.2
Sorry I made a typo at the end: \[\therefore, \binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2} = \binom{2n}{n} ^2\]
 2 years ago

y2o2 Group TitleBest ResponseYou've already chosen the best response.0
yeah , i realized that :)
 2 years ago
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