Question

Give an algebraic proof that (2n choose 2) = 2(n choose 2) + n^2. Hint: The idea is to expand and simplify each side of the question, and then make sure they are both the same. It might help if the two sides are computed for a few specific values. Try n = 3, and n = 4.

I'm lost on this question, please help.

Answer 1

Note that

\(\large \displaystyle {2n\choose 2} = \frac{(2n)!}{2!(2n-2)!} =\ldots\)

and

\(\large\displaystyle 2 {n\choose 2}+n^2=2\cdot \frac{n!}{2!(n-2)!}+n^2 =\ldots \).

Can you take things from here and show that both expressions are equivalent? :-)

Answer 2

bro this can be easily proved by using mathematical induction but it would take a page or two to solve it ,

Answer 3

"Give an \(\Huge\mbox{algebraic}\) proof that (2n choose 2) = 2(n choose 2) + n^2. Hint: The idea is to expand and simplify each side of the question, and then make sure they are both the same. It might help if the two sides are computed for a few specific values. Try n = 3, and n = 4.

I'm lost on this question, please help."

Answer 4

@rottenbologna
Welcome to OpenStudy :D
You may want to read the code of conduct ( http://openstudy.com/code-of-conduct ).
Don't forget to click on the button called "Best Answer" to give a medal to the person who helped you :)

Answer 5

Note that \(\large(2n)!=(2n)\times(2n-1)\times(2n-2)\times\cdots\times(n+2)\times(n+1)\times n!\)

Answer 6

And similarly,
\[\large(2n-2)!=(2n-2)\times(2n-1)\times(2n)\times\cdots\times n\times(n-1)\times(n-2)!\]