If (1+X+x^2)^n = Sigma_(r=0)^(2n)a_(r) x^r then prove that(a) a_(r)=a_(2n-r)(b ) Sigma_(r=0)^(n-1)a_(r)=1/2 (3^n-a_n)

If (1+X+x^2)^n = Sigma_(r=0)^(2n)a_(r) x^r then prove that(a) a_(r)=a_

1.	If (1+X+x^2)^n = Sigma_(r=0)^(2n)a_(r) x^r then prove that(a) a_(r)=a_(2n-r)(b ) Sigma_(r=0)^(n-1)a_(r)=1/2 (3^n-a_n)
Answer» Solution :We have ` (1+X+x^2)^n = UNDERSET(r=0)OVERSET(2n)Sigma a_(r) x^r` Replace x by1/x `therefore (1 +1/x+1/(x^2))^n = underset(r=0)overset(2n) Sigma a_(r)((1)/(x))^r` `rArr(x^2+x+1)^n = underset( r=0)overset(2n)Sigma a_rx^(2n-r)` `underset(r=0)overset(2n ) a_(r)x^r =underset(r=0)overset(2n ) a_(r)x^r "" {"Using (A) "}` Equating the cofficient of `x^(2n-r)` on the both sides ,we get `a_(2n-r)= a_(r) " for " 0 LE r le 2n` Hence`a_(r)=a_(2n-r)` (b) Putting x=1 in given series , then `a_(0)+a_1 +a_2+......+a_(2n)=(1+1+1)^n` `a_(0)+a_(1)+a_(2)+...... +a_(2n)=3^n` But`a^r= a_(2n-r) for 0le r le4 2 n ` `therefore ` Series (1) reduces to `2(a_(0)+a_1+a_2+.......+a_(n-1)) + a_n =3^n` `thereforea_0+a_1+a_2+........+a_(n-1)= 1/2 (3^n-a_n)`