Prove that the greatest value of `.^(2n)C_(r)(0 le r le 2n)` is `.^(2n)C_(n)` (for `1 le r le n)`.

Explanation

We have, `(.^(2n)C_(r))/(.^(2n)C_(r-1))=(2n-r+1)/(r)" "[because(.^(n)C_(r))/(.^(n)C_(r-1))=(n-r+1)/(r)]`
`=(2(n-r)+(r+1))/(r)=(1+2(n-r)+1)/(r)gt1`
`implies(.^(2n)C_(r))/(.^(2n)C_(r-1))gt1" "for" "1lerlen]`
`therefore.^(2n)C_(r-1)lt .^(2n)C_(r)`
On putting `r=1,2,3, . .,n`,
then `.^(2n)C_(0) lt .^(2n)C_(1), .^(2n)C_(2) lt .^(2n)C_(3) lt . . . lt .^(2n)C_(n-1) lt .^(2n)C_(n)`
On combining all inequalities, we get
`implies.^(2n)C_(0) lt .^(2n)C_(1) lt .^(2n)C_(2) lt .^(2n)C_(3) lt . . . lt .^(2n)C_(n-1) lt .^(2n)C_(n)`
but `.^(2n)C_(r)=.^(2n)C_(2n-r)`, it follows that
`.^(2n)C_(2n) lt .^(2n)C_(2n-1) lt .^(2n)C_(2n-2) lt .^(2n)C_(2n-3) lt . . . lt .^(2n)C_(n+1) lt .^(2n)C_(n)`
Hence, the greatest value of `.^(2n)C_(r)` is `.^(2n)C_(n)`.