If `(1+x)^(n) = C_(0)+C_(1)x + C_(2) x^(2) +...+C_(n)x^(n)`
then ` C_(0)""^(2)+C_(1)""^(2) + C_(2)""^(2) +...+C_(n)""^(2)` is equal to

Answer

N/a

Explanation

`(1+x)^(n)=C_(0)+C_(1).x+C_(2).x^(2)+......+C_(n).x^(n) " and " (x+1)^(n)=C_(0).x^(n)+C_(1).x^(n-1)+C_(2).x^(n-2)+......+C_(n)`
Multiple both eqs. We get .
`(1+x)^(2n)=(C_(0)+C_(1).x+C_(2).x^(2)+....+C_(n).x^(n))`
`(C_(0).x^(n)+C_(1).x^(n-1)+C_(2).x^(n-2)+....+C_(n))`
Now
`C_(0)^(2)+C_(1)^(2)+C_(2)^(2)+....+C_(n)^(2)=` Coefficient of `x^(n)` in the expansion of `(1+x)^(2n)`
`T_(r+1)=^(2n)C_(r).(1)^(2n-r).x^(r)=^(2n)C_(r).x^(r)`
`:." Coefficient of " x^(n)=^(2n)C_(n)=(|ul2n)/(|uln|uln)`
Therefore
`C_(0)^(2)+C_(1)^(2)+C_(2)^(2)+.....+C_(n)^(2)=(|ul2n)/(|uln|uln)` Hence Proved.