[" If "(1+x+x^(2))^(n)=sum_(r=0)^(2n)a_(r)x'," then "a_(1)-2a_(2)+3a_(3)-...-2n*a_(2n)=...],[[" 1) "0," 2) "1," 3) "n," 4) "-n]]

If (1+x+x^(2))^(n)=sum_(r=0)^(2n)a_(r)x^(r), then a_(1)-2a_(2)+3a_(3)-...-2na_(2n)=

If (1+x+x^(2))^(n) = sum_(r=0)^(2 n) a_(r). x^(r) then a_(1)-2 a_(2)+3 a_(3)-..-2 n. a_(2n) =

If (1+2x+x^(2))^(n) = sum_(r=0)^(2n)a_(r)x^(r) , then a_(r) =

If (1+2x+x^(2))^(n) = sum_(r=0)^(2n)a_(r)x^(r) , then a_(r) =

If (1+2 x+x^(2))^(n)=sum_(r=0)^(2 n) a_(r) x^(r) , then a_(r) =

If (1+x+x^(2))^(n)=sum_(r=0)^(2n)a_(r)x^(r), then prove that a_(r)=a_(2n-r)

If (n) is an odd positive integer and (1+x+x^(2)+x^(3))^(n) = sum_(r=0)^(3 n) a_(r) x^(r) then, a_(0)-a_(1)+a_(2)-a_(3)+.. .-a_(3 n) =

If (1+x)^(2n)=sum_(r=0)^(2n)a_(r)x^(r) then the value of a_(0)^(2)-a_(1)^(2)+a_(2)^(2)-a_(3)^(2)+...+(a_(2n))^(2)=

Consider (1 + x + x^(2))^(n) = sum_(r=0)^(2n) a_(r) x^(r) , where a_(0), a_(1), a_(2),…, a_(2n) are real number and n is positive integer. If n is even, the value of sum_(r=0)^(n/2-1) a_(2r) is