If `A=[1 1 1 1 1 1 1 1 1]`, prove that `A^n=[3^(n-1)3^(n-1)3^(n-1)3^(n-1)3^(n-1)3^(n-1)3^(n-1)3^(n-1)3^(n-1)], n in Ndot`

If A=[111111111], then prove that A^(n)=[3^(n-1)3^(n-1)3^(n-1)3^(n-1)3^(n-1)3^(n-1)3^(n-1)3^(n-1)3^(n-1)] for every positive integer n

If A,=[[1,1,11,1,11,1,1]]A^(n)=,[[3^(n-1),1]]3^(n-1),3^(n-1),3^(n-1)3^(n-1),3^(n-1),3^(n-1)]]

Prove that 3^(n+1)gt3(n+1)

(2.3^(n+1)+7.3^(n-1))/(3^(n+1)-2((1)/(3))^(1-n))=

Show that (n-1)/(n+1)+3((n-1)/(n+1))^2+5((n-1)/(n+1))^3+....+oo=sum_(r=1)^(n-1) r .

Prove that ((2n+1)!)/(n!)=2^(n){1.3.5(2n-1)(2n+1)}

Simplify: (3^(n+1))/(3^(n(n-1)))-:(9^(n+1))/((3^(n+1))^((n-1)))

Prove that ((2n+1)!)/(n!)=2^(n)[1.3.5.....(2n-1)*(2n+1)]

1^(3)+2^(3)+3^(3)+.....+n^(3)=(n(n+1)^(2))/(4), n in N

Prove that 1^(1)xx2^(2)xx3^(3)xx xx n^(n)<=[(2n+1)/3]^(n(n+1)/2),n in N