Prove that `(""^(2n)C_(0))^(2)-(""^(2n)C_(1))^(2)+(""^(2n)C_(2))^(2)-…+(""^(2n)C_(2n))^(2)=(-1)^(n)*""^(2n)C_(n)`.

Explanation

Since , `(1-x)^(2n) = ""^(2n)C_(0) - ""^(2n)C_(1)x + ""^(2n)C_(2) x^(2) - ...+ (-1)^(2n) *""^(2n)C_(2n) x^(2n)`
or `(1-x)^(2n) = ""^(2n)C_(0) - ""^(2n)C_(1)x + ""^(2n)C_(2) x^(2) - ...+ ""^(2n)C_(2n) x^(2n)` ...(i)
and ` (x +1)^(2n) = ""^(2n)C_(0) x^(2n) + ""^(2n)C_(1) + x^(2n-1) + ...+ ""^(2n)C_(2n) `... (ii)
`(x^(2) -1)^(2n) = (""^(2n)C_(0)- ""^(2n)C_(1)x + ""^(2n)C_(2)x^(2) - ...+ ""^(2n)C_(2n) x^(2n))`
`xx(""^(2n)C_(0)x^(2n) + ""^(2n)C_(1)x^(2n-1)+""^(2n)C_(2)x^(2n-2) + ...+ ""^(2n)C_(2n) )` ...(iii)
Now coefficient of `x^(2n)` in RHS
` = (""^(2n)C_(0))^(2) - (""^(2n)C_(1))^(2) + (""^(2n)C_(2))^(2) - ...+ (""^(2n)C_(2n))^(2)`
Now ,LHS can also be written as `(1 + x^(2))^(2n)`.
` therefore ` General term in LHS , ` T_(r+1) = ""^(2n)C_(r) (-x^(2))^(r)`
Putting r = n , we get ` T_(n+1) = (-1)^(n) * ""^(2n)C_(n) x^(2n)`
` rArr " Coefficient of " x^(2n) ` in LHS ` = (-1)^(n) *""^(2n)C_(n)`
But Eq.(iii) is an identity , therefore coefficient of ` x^(2n)` in
RHS = coefficient of ` x^(2n)` LHS
`rArr (""^(2n)C_(0))^(2) - (""^(2n)C_(1))^(2) + (""^(2n)C_(2))^(2) - ...+ (""^(2n)C_(2n))^(2n)`
` = (-1)^(n)*""^(2n)C_(n)`
Aliter
Since , `(1 + x)^(2n) + ""^(2n)C_(0) + ""^(2n)C_(1)x + ""^(2n)C_(2)x^(2) + ...+ ""^(2n)C_(2n) x^(2n) ...`(i)
and ` (1 - (1)/(x))^(2n) = ""^(2n)C_(0) = (""^(2n)C_(1))/(x) + (""^(2n)C_(2))/(x^(2)) -...+ (""^(2n)C_(2n))/(x^(2n))` ...(iii)
On multiplying Eqs .(i) and (ii) , we get
`((x^(2) -1)^(2n))/(x^(2n)) = (""^(2n)C_(0) + ""^(2n)C_(1)x + ""^(2n)C_(2)x^(2) + ...+ ""^(2n)C_(2n)x^(2n))`
` xx(""^(2n)C_(0) - (""^(2n)C_(1))/(x) +( ""^(2n)C_(2))/(x^(2))-...+(""^(2n)C_(2n))/(x^(2n)))`...(iii)
Now , constant term in RHS
`= (""^(2n)C_(0))^(2) - (""^(2n)C_(1))^(2)+ (""^(2n)C_(2))^(2)- ...+ (""^(2n)C_(2n))^(2)`
Constant term in LHS = Constant term in `((x^(2) -1)^(2n))/(x^(2n))`
= Coefficient of ` x^(2n) " in" (x^(2) -1)^(2n)`
= Coefficient of ` x^(2n)" in " (1 - x^(2)) ^(2n)`
` = ""^(2n)C_(n) (-1)^(n)= (-1)^(n) *""^(2n)C_(n)`
But Eq.(iii) is an identity , therefore the constant term in
RHS = constant term in LHS .
` rArr (""^(2n)C_(0))^(2) - (""^(2n)C_(1))^(2) + (""^(2n)C_(2))^(2)- ...+ (""^(2n)C_(2n))^(2)`
` = (-1)^(n) *""^(2n)C_(n)` .