((2^(n)+2^(n-1)))/((2^(n+1)-2^(n)))

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

If A = [(1,1),(1,1)] , prove by induction that A^n = [(2^(n-1), 2^(n-1)), (2^(n-1), 2^(n-1))] for all natural numbers n.

If A=[[1,1],[1,1]] ,prove that A^n=[[2^(n-1),2^(n-1)],[2^(n-1),2^(n-1)]], for all positive integers n.

If A=[[1,1],[1,1]] ,prove that A^n=[[2^(n-1),2^(n-1)],[2^(n-1),2^(n-1)]] , for all positive integers n.

[ If 2 is the sum of infinity of a G.P.,whose first clement is 1 ,then the sum of the first n terms is [ 1) (2^(n)-1)/(2^(n)), 2) (2^(n)-1)/(2^(n-1)), 3) (2^(n-1)-2)/(2), 4) (2^(n-1)-1)/(2^(n))]]

Divide x^(2n)+a^(2^(n-1))x^(2^(n-1))+a^(2^(n))byx^(2^(n-1))-a^(2^(n-2))x^(2^(n-2))+a^(2^(n-1))

If n is a positive integer, prove that 1-2n+(2n(2n-1))/(2!)-(2n(2n-1)(2n-2))/(3!)+.......+(-1)^(n-1)(2n(2n-1)(n+2))/((n-1)!)= (-1)^(n+1)(2n)!//2(n !)^2dot

If n is a positive integer,prove that 1-2n+(2n(2n-1))/(2!)-(2n(2n-1)(2n-2))/(3!)+...+(-1)^(n-1)(2n(2n-1)...(n+2))/((n-1)!)=(-1)^(n+1)(2n)!/2(n!)^(2)

If n is a positive integer, prove that 1-2n +(2n(2n-1))/(2!) - (2n(2n-1) (2n-2))/(3!) +… + (-1)^(n-1) (2n(2n-1) …(n+2))/((n-1)!) = (-1)^(n+1) ((2n)!)/(2(n!)^(2))

If n is a positive integer, prove that 1-2n+(2n(2n-1))/(2!)-(2n(2n-1)(2n-2))/(3!)++(-1)^(n-1)(2n(2n-1)(n+2))/((n-1)!)=(-1)^(n+1)(2n)!//2(n !)^2dot