2^(n)-2^(n-1)

show that (3*2^(n+1)+2^(n))/(2^(n+2)-2^(n-1))=2

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.

Show : 2^(n)-(n)/(1!).2^(n-1)+(n(n-1))/(2!).2^(n-2)-....+(-1)^(n)=1

FInd lim_(n rarr oo)(2n-1)2^(n)(2n+1)^(-1)2^(1-n)

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.

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=2^(n-1).(2^n-1) where 2^n-1 is a prime, then the sum of the all divisors of N is

Find the sum of the series: 1. n+2.(n-1)+3.(n-2)++(n-1). 2+n .1.