## y2o2 3 years ago prove that :

1. y2o2

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

2. KingGeorge

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

3. shubhamsrg

you can use induction..

4. Zarkon

Use Vandermonde's identity

5. Mimi_x3

well have you tried it?

6. Mimi_x3

since it was from 2 months ago..

7. y2o2

Yes , and it came up with nothing

no it works.

9. Mimi_x3

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

11. Mimi_x3

whats wrong with the method i used?

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

13. Mimi_x3

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

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

15. y2o2

thank you, i got it now :)

16. Mimi_x3

You're welcome (:

17. Mimi_x3

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

18. y2o2

yeah , i realized that :)