Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

prove that :

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

\[\large {^{2n} C _n} = {(^n C _0)}^2 +{(^n C _1)}^2 +{(^n C _2)}^2 +...+{(^n C _n)}^2 \]
So in summation notation, show that\[\binom{2n}{n}=\sum_{i=0}^n \binom{n}{i}^2\]
you can use induction..

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Use Vandermonde's identity
well have you tried it?
since it was from 2 months ago..
Yes , and it came up with nothing
no it works.
\[LHS: \binom{2n}{n} x^{n} \] \[RHS\left[\binom{n}{0}\binom{n}{n}+\binom{n}{1}\binom{n}{n-1}+\binom{n}{2}\binom{n}{n-2}+....+\binom{n}{n-2}\binom{n}{2}+\binom{n}{n-1}\binom{n}{1}+\binom{n}{n}\binom{n}{0}\right] *x^n\] \[By ~Symmetry: \binom{n}{n} = \binom{n}{0} ; \binom{n}{n-1} = \binom{n}{1} \] \[hence~ RHS = \left[\binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2}\right] x^n\] \[hence, coeeficients~ of~ x^n~ are~ the ~same \] \[\therefore, \binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2} = \binom{2n}{n} \]
use this. its better.
whats wrong with the method i used?
nothing. I said its easier to use vandermonde's identity.
okay; well i dont know how to use that identity lol
just plug in n every where except at k(its a summation variable).
thank you, i got it now :)
You're welcome (:
Sorry I made a typo at the end: \[\therefore, \binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2} = \binom{2n}{n} ^2\]
yeah , i realized that :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question