If `(1+x+x^2)^n = underset(2n)overset(r=0)Sigma a _(r)x^r` then prove that
(a) `a_(r)= 20_(2n-r)`
(b) `underset(r=0)overset(2 n)Sigma a_(r) = 1/2 (3^n-a_(n))`

Explanation

We have
` (1+x+x^2)^n = underset(r=0)overset(2n)Sigma a_(r) x^r`
Replace x by 1/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 0 le r le4 2 n `
`therefore ` Series (1) reduces to
`2(a_(0)+a_1+a_2+.......+a_(n-1)) + a_n =3^n`
`therefore a_0+a_1+a_2+........+a_(n-1)= 1/2 (3^n-a_n)`